ML Aggarwal Class 10 Maths Solutions Chapter 09 Arithmetic and Geometric Progression

Exercise 9.1

Question 1. For the following A.P.s, write the first term a and the common difference d.
(i) 3, 1, -1, -3, ....
(ii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3} \)
(iii) -3.2, -3, -2.8, -2.6, ...
Answer: (i) The sequence starts with 3. Subtracting consecutive terms shows the common difference is -2. So a = 3 and d = -2.
(ii) Starting term is \( \frac{1}{3} \). The common difference can be found by taking the second term minus the first term: \( \frac{5}{3} - \frac{1}{3} = \frac{4}{3} \). Thus a = \( \frac{1}{3} \) and d = \( \frac{4}{3} \).
(iii) The opening term is -3.2. Subtracting gives -3 - (-3.2) = 0.2. Therefore a = -3.2 and d = 0.2.
In simple words: The first term is where the sequence begins. The common difference is the amount we add each time to get the next number.

Exam Tip: Always verify the common difference by checking that it's the same between any two consecutive terms. A single calculation error can make the difference invalid.

Question 2. Write first four of the terms of the A.P., when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = \( \frac{1}{2} \), d = \( -\frac{1}{6} \)
Answer: (i) Starting with a = 10, we add d = 10 to each term. The first term is 10. Add 10 to get 20. Add 10 again to get 30. Add once more to get 40. The sequence is 10, 20, 30, 40.
(ii) Starting with a = -2, we add d = 0 to each term. Since we're adding zero, every term stays the same: -2, -2, -2, -2.
(iii) Starting with a = 4, we add d = -3 to each term. From 4, subtract 3 to get 1. From 1, subtract 3 to get -2. From -2, subtract 3 to get -5. The sequence is 4, 1, -2, -5.
(iv) Starting with a = \( \frac{1}{2} \), we add d = \( -\frac{1}{6} \) to each term. The second term: \( \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \). The third term: \( \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \). The fourth term: \( \frac{1}{6} - \frac{1}{6} = 0 \). The sequence is \( \frac{1}{2}, \frac{1}{3}, \frac{1}{6}, 0 \).
In simple words: Multiply the common difference by position and add it to the first term. Keep adding d over and over to build the list.

Exam Tip: For fractional common differences, always find a common denominator before performing arithmetic to avoid careless mistakes.

Question 3. Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms:
(i) 4, 10, 16, 22, ....
(ii) -2, 2, -2, 2, ....
(iii) 2, 4, 8, 16, ....
(iv) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \)
(v) -10, -6, -2, 2, ...
(vi) 1², 3², 5², 7², ....
Answer: (i) Check the differences: 10 - 4 = 6, 16 - 10 = 6, 22 - 16 = 6. The difference is always 6 (fixed). This is an A.P. with d = 6. The next three terms are found by adding 6 each time: 28, 34, 40.
(ii) Check the differences: 2 - (-2) = 4, but -2 - 2 = -4. The differences are not the same. This is not an A.P.
(iii) Check the differences: 4 - 2 = 2, 8 - 4 = 4, 16 - 8 = 8. The differences keep changing. This is not an A.P.
(iv) Check the differences: \( \frac{5}{2} - 2 = \frac{1}{2} \), \( 3 - \frac{5}{2} = \frac{1}{2} \), \( \frac{7}{2} - 3 = \frac{1}{2} \). The difference is constant at \( \frac{1}{2} \) (fixed). This is an A.P. with d = \( \frac{1}{2} \). The next three terms are 4, \( \frac{9}{2} \), 5.
(v) Check the differences: -6 - (-10) = 4, -2 - (-6) = 4, 2 - (-2) = 4. The difference is always 4 (fixed). This is an A.P. with d = 4. The next three terms are 6, 10, 14.
(vi) First rewrite as: 1, 9, 25, 49, .... Check the differences: 9 - 1 = 8, 25 - 9 = 16, 49 - 25 = 24. The differences are not constant. This is not an A.P.
In simple words: An A.P. happens when you add the same number each time. Check if all differences match. If they do, it's an A.P.

Exam Tip: Calculate at least three consecutive differences to confirm the sequence is (or is not) an A.P. One matching difference is not enough.

Exercise 9.2

Question 1. Find the A.P. whose nth term is 7 - 3n. Also find the 20th term.
Answer: We're given the formula \( a_n = 7 - 3n \). Plug in n = 1, 2, 3, 4 to find the sequence. When n = 1: \( a_1 = 7 - 3(1) = 4 \). When n = 2: \( a_2 = 7 - 3(2) = 1 \). When n = 3: \( a_3 = 7 - 3(3) = -2 \). When n = 4: \( a_4 = 7 - 3(4) = -5 \). So the sequence is 4, 1, -2, -5, .... To find the 20th term, substitute n = 20: \( a_{20} = 7 - 3(20) = 7 - 60 = -53 \). The A.P. is 4, 1, -2, -5, ... and the 20th term is -53.
In simple words: Replace n with each number in the formula. The answers give you the sequence. Plug in 20 to get the 20th term.

Exam Tip: Always substitute the values correctly into the given formula. A careless sign error when multiplying can flip the answer.

Question 2. Find the indicated terms in each of the following A.P.s:
(i) 1, 6, 11, 16, ....; \( a_{20} \)
(ii) -4, -7, -10, -13, ...., \( a_{25} \), \( a_n \)
Answer: (i) The sequence is 1, 6, 11, 16, .... Subtracting shows 6 - 1 = 5, so d = 5 and a = 1. The nth term formula is \( a_n = a + (n - 1)d = 1 + (n - 1) \cdot 5 = 1 + 5n - 5 = 5n - 4 \). When n = 20: \( a_{20} = 5(20) - 4 = 100 - 4 = 96 \). So \( a_{20} = 96 \).
(ii) The sequence is -4, -7, -10, -13, .... Subtracting shows -7 - (-4) = -3, so d = -3 and a = -4. The nth term formula is \( a_n = a + (n - 1)d = -4 + (n - 1)(-3) = -4 - 3n + 3 = -3n - 1 \). When n = 25: \( a_{25} = -3(25) - 1 = -75 - 1 = -76 \). The general nth term is \( a_n = -3n - 1 \). So \( a_{25} = -76 \) and \( a_n = -3n - 1 \).
In simple words: Find the first term and the common difference. Use the formula \( a_n = a + (n - 1)d \). Plug in the value of n you need.

Exam Tip: When you derive the general formula for \( a_n \), double-check it by substituting n = 1 to verify you get the first term.

Question 3. Find the nth term and the 12th term of the list of numbers: 5, 2, -1, -4, ...
Answer: The sequence is 5, 2, -1, -4, .... Subtracting shows 2 - 5 = -3, so d = -3 and a = 5. Using \( a_n = a + (n - 1)d \), we get \( a_n = 5 + (n - 1)(-3) = 5 - 3n + 3 = 8 - 3n \). When n = 12: \( a_{12} = 8 - 3(12) = 8 - 36 = -28 \). Therefore \( a_{12} = -28 \) and the general term is \( a_n = 8 - 3n \).
In simple words: Find d by subtracting any two next-door terms. Write the formula using a and d. Put 12 in place of n to get the 12th term.

Exam Tip: Verify your formula by checking it works for the first few terms before using it to find distant terms.

Question 4(i). If the common difference of an A.P. is -3 and the 18th term is -5, then find its first term.
Answer: We know \( a_n = a + (n - 1)d \). Given: d = -3 and \( a_{18} = -5 \). Substitute n = 18 into the formula: \( -5 = a + (18 - 1)(-3) = a + 17 \times (-3) = a - 51 \). Solving for a: \( a = -5 + 51 = 46 \). The first term is 46.
In simple words: Put the values you know into the formula. Solve for the unknown (the first term) using basic algebra.

Exam Tip: Be careful with negative signs. When d is negative and you multiply by (n - 1), make sure your arithmetic stays correct.

Question 4(ii). If the first term of an A.P. is -18 and its 10th term is zero, then find its common difference.
Answer: We know \( a_n = a + (n - 1)d \). Given: a = -18 and \( a_{10} = 0 \). Substitute n = 10: \( 0 = -18 + (10 - 1)d = -18 + 9d \). Solving for d: \( 9d = 18 \), so \( d = 2 \). The common difference is 2.
In simple words: Plug what you know into the formula. Use algebra to find what's missing.

Exam Tip: When an equation has a known term equal to zero, isolate the variable term first, then divide.

Question 5. Which term of the A.P.
(i) 3, 8, 13, 18, ... is 78?
(ii) \( 18, 15\frac{1}{2}, 13, \ldots \) is -47?
Answer: (i) The sequence is 3, 8, 13, 18, .... We have a = 3 and d = 8 - 3 = 5. We want to find n such that \( a_n = 78 \). Using \( a_n = a + (n - 1)d \): \( 78 = 3 + (n - 1) \cdot 5 = 3 + 5n - 5 = 5n - 2 \). So \( 5n = 80 \), giving \( n = 16 \). Thus 78 is the 16th term.
(ii) The sequence is \( 18, 15\frac{1}{2}, 13, \ldots \), which can be written as \( 18, \frac{31}{2}, 13, \ldots \). We have a = 18 and \( d = \frac{31}{2} - 18 = \frac{31 - 36}{2} = -\frac{5}{2} \). We want n such that \( a_n = -47 \). Using the formula: \( -47 = 18 + (n - 1) \cdot (-\frac{5}{2}) = 18 - \frac{5(n - 1)}{2} = 18 - \frac{5n - 5}{2} \). Multiply by 2: \( -94 = 36 - 5n + 5 = 41 - 5n \). So \( 5n = 41 + 94 = 135 \), giving \( n = 27 \). Thus -47 is the 27th term.
In simple words: Set the term equal to the value you're looking for. Solve for n. That n is which term it is.

Exam Tip: With fractional common differences, carefully manage the arithmetic by converting to common denominators or working step by step.

Question 6(i). Check whether -150 is a term of the A.P. 11, 8, 5, 2, ...
Answer: The sequence is 11, 8, 5, 2, .... We have a = 11 and d = 8 - 11 = -3. We check if there exists a natural number n such that \( a_n = -150 \). Using \( a_n = a + (n - 1)d \): \( -150 = 11 + (n - 1)(-3) = 11 - 3n + 3 = 14 - 3n \). So \( 3n = 14 + 150 = 164 \), giving \( n = \frac{164}{3} \). Since n is not a whole number, -150 is not a term of this A.P.
In simple words: Use the formula to see what n would have to be. If n comes out to a whole number, it's a term. If not, it isn't.

Exam Tip: A term belongs to the sequence only if n is a positive integer. Fractional or negative values of n mean the number is not in the sequence.

Question 6(ii). Find whether 55 is a term of the A.P. 7, 10, 13, .... or not. If yes, find which term is it.
Answer: The sequence is 7, 10, 13, .... We have a = 7 and d = 10 - 7 = 3. We check if there exists a natural number n such that \( a_n = 55 \). Using \( a_n = a + (n - 1)d \): \( 55 = 7 + (n - 1) \cdot 3 = 7 + 3n - 3 = 4 + 3n \). So \( 3n = 51 \), giving \( n = 17 \). Since n is a positive whole number, 55 is a term of the A.P., and it is the 17th term.
In simple words: Set up the equation. Solve for n. If you get a whole number, that's the position.

Exam Tip: Always verify your answer by substituting the found n back into the formula to confirm you get the target value.

Question 7(i). Find the 20th term from the last term of the A.P. 3, 8, 13, ..., 253.
Answer: We have d = 8 - 3 = 5 and the last term l = 253. The formula for the nth term from the end is: nth term from end = l - (n - 1)d. So the 20th term from the end = \( 253 - (20 - 1) \cdot 5 = 253 - 19 \times 5 = 253 - 95 = 158 \). The 20th term from the last is 158.
In simple words: Start from the last number. Go backward by counting off the common difference again and again.

Exam Tip: The formula for counting backward from the end uses the same common difference but subtracts instead of adds.

Question 7(ii). Find the 12th term from the end of A.P. -2, -4, -6, ...., -100.
Answer: The common difference is d = -4 - (-2) = -2, and the last term is l = -100. Using the formula for the nth term from the end: nth term from end = l - (n - 1)d. Substituting the values: 12th term from end = -100 - (12 - 1)(-2) = -100 - 11 × (-2) = -100 + 22 = -78.
In simple words: Start at the last number -100. Move backward 11 steps, where each step is the opposite of the common difference. You get -78.

Exam Tip: When finding terms from the end, remember the common difference changes sign in the formula - if d is negative, -d becomes positive, which adds to your starting value.

Question 8. Find the sum of the two middle most terms of the A.P. -\( \frac{4}{3} \), -1, -\( \frac{2}{3} \), ...., \( 4\frac{1}{3} \).
Answer: Given: a = -\( \frac{4}{3} \), d = -1 - (-\( \frac{4}{3} \)) = \( \frac{1}{3} \), and l = \( 4\frac{1}{3} \) = \( \frac{13}{3} \).

To find the number of terms, use the formula: \( a_n = a + (n - 1)d \)

\( \frac{13}{3} = -\frac{4}{3} + (n - 1)\frac{1}{3} \)

\( \frac{13}{3} + \frac{4}{3} = \frac{(n - 1)}{3} \)

\( \frac{17}{3} = \frac{(n - 1)}{3} \)

Multiplying both sides by 3: 17 = n - 1, so n = 18.

Since there are 18 terms, the two middle terms are the 9th and 10th terms.

\( a_9 = a + 8d = -\frac{4}{3} + 8 × \frac{1}{3} = -\frac{4}{3} + \frac{8}{3} = \frac{4}{3} \)

\( a_{10} = a + 9d = -\frac{4}{3} + 9 × \frac{1}{3} = -\frac{4}{3} + \frac{9}{3} = \frac{5}{3} \)

Sum of the two middle most terms = \( a_9 + a_{10} = 2a + 17d = 2 × (-\frac{4}{3}) + 17 × \frac{1}{3} = -\frac{8}{3} + \frac{17}{3} = \frac{9}{3} = 3 \).
In simple words: Find how many terms are in the sequence. The middle two terms are the 9th and 10th. Add them together to get 3.

Exam Tip: When finding middle terms in an A.P., the sum of the middle two terms for an even number of terms always equals the sum of the first and last terms.

Question 9. Which term of the A.P. 53, 48, 43, .... is the first negative term?
Answer: First, identify which term equals zero, since the next term will be the first negative one. Here: a = 53, d = 48 - 53 = -5, \( a_n \) = 0. Using \( a_n = a + (n - 1)d \):

\( a_n = 53 + (n - 1)(-5) \)

\( 0 = 53 - 5n + 5 \)

\( 5n = 58 \)

\( n = 11\frac{3}{5} \)

Since n is not a whole number, no term equals exactly zero. The next whole number is 12. Therefore, the 12th term is the first negative term.
In simple words: Work backward from when the sequence hits zero. The term number that gives zero is between 11 and 12. So the 12th term is the first one that is negative.

Exam Tip: When a term number is not a whole number, always round up to the next integer to find which position actually contains a negative value.

Question 10. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer: Given: \( a_3 \) = 16 (Eq 1) and \( a_7 \) - \( a_5 \) = 12 (Eq 2).

From Eq 1 using \( a_n = a + (n - 1)d \):

\( a + 2d = 16 \)

\( a = 16 - 2d \) (Eq 3)

From Eq 2:

\( a_7 - a_5 = 12 \)

\( [a + 6d] - [a + 4d] = 12 \)

\( 2d = 12 \)

\( d = 6 \)

Substituting into Eq 3: \( a = 16 - 2(6) = 16 - 12 = 4 \).

The sequence is: \( a_1 = 4 \), \( a_2 = 10 \), \( a_3 = 16 \), \( a_4 = 22 \), ....
In simple words: Set up two equations using the given information. Solve them to find the first term is 4 and the common difference is 6.

Exam Tip: Always write out at least the first four terms to verify your answer is correct - check that the 3rd term is 16 and that the difference between the 7th and 5th is 12.

Question 11. Find the 20th term of the A.P. whose 7th term is 24 less than the 11th term, first term being 12.
Answer: Given: a = 12 and \( a_{11} \) - \( a_7 \) = 24 (Eq 1).

Using \( a_n = a + (n - 1)d \) for Eq 1:

\( a_{11} - a_7 = 24 \)

\( [a + 10d] - [a + 6d] = 24 \)

\( 4d = 24 \)

\( d = 6 \)

Finding the 20th term:

\( a_{20} = 12 + (20 - 1) × 6 = 12 + 19 × 6 = 12 + 114 = 126 \)
In simple words: Use the relationship between two terms to find the common difference. Then apply the formula to locate the 20th term.

Exam Tip: When you know the first term explicitly, always substitute it into formulas directly rather than solving for it - this saves time and reduces calculation errors.

Question 12. Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73.
Answer: Given: \( a_{11} \) = 38 (Eq 1) and \( a_6 \) = 73 (Eq 2).

Using \( a_n = a + (n - 1)d \) for Eq 1:

\( a + 10d = 38 \)

\( a = 38 - 10d \) (Eq 3)

Using the formula for Eq 2:

\( a + 5d = 73 \) (Eq 4)

Substituting Eq 3 into Eq 4:

\( 38 - 10d + 5d = 73 \)

\( 38 - 5d = 73 \)

\( -5d = 35 \)

\( d = -7 \)

Therefore: \( a = 38 - 10(-7) = 38 + 70 = 108 \).

The 31st term: \( a_{31} = 108 + 30(-7) = 108 - 210 = -102 \)
In simple words: You have two pieces of information about two different terms. Create two equations and solve the system to find a and d, then calculate the requested term.

Exam Tip: When solving a system with two unknowns (a and d), express one in terms of the other from the first equation, then substitute into the second to isolate the common difference.

Question 13. If the seventh term of an A.P. is \( \frac{1}{9} \) and its ninth term is \( \frac{1}{7} \), find its 63rd term.
Answer: Given: \( a_7 = \frac{1}{9} \) (Eq 1) and \( a_9 = \frac{1}{7} \) (Eq 2).

Using \( a_n = a + (n - 1)d \) for Eq 1:

\( a + 6d = \frac{1}{9} \)

Multiplying by 9: \( 9a + 54d = 1 \)

\( a = \frac{1 - 54d}{9} \) (Eq 3)

Using the formula for Eq 2:

\( a + 8d = \frac{1}{7} \) (Eq 4)

Substituting Eq 3 into Eq 4:

\( \frac{1 - 54d}{9} + 8d = \frac{1}{7} \)

\( \frac{1 - 54d + 72d}{9} = \frac{1}{7} \)

\( \frac{1 + 18d}{9} = \frac{1}{7} \)

\( 7(1 + 18d) = 9 \)

\( 7 + 126d = 9 \)

\( 126d = 2 \)

\( d = \frac{1}{63} \)

Therefore: \( a = \frac{1 - 54 × \frac{1}{63}}{9} = \frac{1 - \frac{54}{63}}{9} = \frac{\frac{63 - 54}{63}}{9} = \frac{\frac{9}{63}}{9} = \frac{9}{9 × 63} = \frac{1}{63} \)

The 63rd term: \( a_{63} = a + 62d = \frac{1}{63} + 62 × \frac{1}{63} = \frac{1 + 62}{63} = \frac{63}{63} = 1 \)
In simple words: Set up two equations with the two given terms. Solve to find a and d. Then calculate the 63rd term using the formula.

Exam Tip: With fractional terms, multiply through by the denominators early to clear fractions and make the algebra cleaner - this reduces the chance of errors in subsequent steps.

Question 14(i). The 15th term of an A.P. is 3 more than twice its 7th term. If the 10th term of the A.P. is 41, find its nth term.
Answer: Given: \( a_{10} \) = 41 (Eq 1) and \( a_{15} \) = 2\( a_7 \) + 3 (Eq 2).

From Eq 1 using \( a_n = a + (n - 1)d \):

\( a + 9d = 41 \)

\( a = 41 - 9d \) (Eq 3)

From Eq 2:

\( a + 14d = 2(a + 6d) + 3 \)

\( a + 14d = 2a + 12d + 3 \)

Substituting Eq 3:

\( 41 - 9d + 14d = 2(41 - 9d) + 12d + 3 \)

\( 41 + 5d = 82 - 18d + 12d + 3 \)

\( 41 + 5d = 85 - 6d \)

\( 11d = 44 \)

\( d = 4 \)

Therefore: \( a = 41 - 9(4) = 41 - 36 = 5 \).

The nth term: \( a_n = 5 + 4(n - 1) = 5 + 4n - 4 = 4n + 1 \)
In simple words: Use both conditions to write two separate equations. Substitute one into the other to solve for d, then find a. Finally, write the general formula.

Exam Tip: Always verify your final answer by checking that both original conditions are satisfied - substitute the values of a and d back to confirm the 10th term is 41 and the relationship between the 15th and 7th terms holds.

Question 14(ii). The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P.
Answer: Given: \( a_5 \) + \( a_7 \) = 52 (Eq 1) and \( a_{10} \) = 46 (Eq 2).

Using \( a_n = a + (n - 1)d \) for Eq 1:

\( [a + 4d] + [a + 6d] = 52 \)

\( 2a + 10d = 52 \)

\( a + 5d = 26 \)

\( a = 26 - 5d \) (Eq 3)

From Eq 2:

\( a + 9d = 46 \)

Substituting Eq 3:

\( 26 - 5d + 9d = 46 \)

\( 26 + 4d = 46 \)

\( 4d = 20 \)

\( d = 5 \)

Therefore: \( a = 26 - 5(5) = 26 - 25 = 1 \).

The A.P. is: 1, 6, 11, 16, 21, ....
In simple words: Add the 5th and 7th terms together. This gives you one equation. Combine with the 10th term condition to find a and d. Write out the first few terms.

Exam Tip: When asked to "find the A.P.", always write out the first few terms explicitly - this shows you understand what an arithmetic progression is and makes it easy to verify your answer.

Question 15. If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term.
Answer: Given: \( a_8 \) = 0.

Using \( a_n = a + (n - 1)d \):

\( a + 7d = 0 \)

\( a = -7d \)

Finding the 38th term:

\( a_{38} = a + 37d = -7d + 37d = 30d \)

Finding the 18th term:

\( a_{18} = a + 17d = -7d + 17d = 10d \)

Comparing: \( a_{38} = 30d = 3 × 10d = 3 × a_{18} \)

Therefore, the 38th term is triple the 18th term. Hence proved.
In simple words: Since the 8th term is zero, you can express a in terms of d. Use this to find the 38th and 18th terms, then show one is three times the other.

Exam Tip: In proof questions, clearly state what you need to prove at the start. Show each step logically, and always end with "Hence proved" - examiners look for this structural clarity.

Question 16. Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13th term?
Answer: Here: common difference d = 10 - 3 = 7 and a = 3. Let the nth term be 84 more than the 13th term:

\( a_n - a_{13} = 84 \)

\( [a + (n - 1)d] - [a + 12d] = 84 \)

\( 3 + 7(n - 1) - (3 + 12 × 7) = 84 \)

\( 3 + 7n - 7 - (3 + 84) = 84 \)

\( 7n - 4 - 87 = 84 \)

\( 7n - 91 = 84 \)

\( 7n = 175 \)

\( n = 25 \)
In simple words: Set up an equation where one term minus another term equals 84. Solve for n.

Exam Tip: Double-check by calculating both the 25th and 13th terms: \( a_{25} = 3 + 24(7) = 171 \) and \( a_{13} = 3 + 12(7) = 87 \); the difference is 171 - 87 = 84. ✓

Question 17(i). How many two digits numbers are divisible by 3?
Answer: Two-digit numbers divisible by 3 form an A.P.: 12, 15, 18, 21, 24, ......, 99. Here: first number a = 12, common difference d = 15 - 12 = 3, and last number = 99.

Using \( a_n = a + (n - 1)d \):

\( 99 = 12 + 3(n - 1) \)

\( 99 = 12 + 3n - 3 \)

\( 99 = 9 + 3n \)

\( 90 = 3n \)

\( n = 30 \)
In simple words: Find the first and last two-digit numbers divisible by 3. They form a sequence that increases by 3 each time. Count how many numbers are in this sequence.

Exam Tip: Always identify the first and last terms carefully - the smallest two-digit number is 10 (not divisible by 3), so start at 12; the largest is 99 (which is divisible by 3).

Question 17(ii). Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer: Numbers divisible by both 2 and 5 are divisible by 10 (since LCM of 2 and 5 is 10). These are: 110, 120, 130, 140, ...., 990. Here: first number a = 110, common difference d = 120 - 110 = 10, and last number = 990.

Using \( a_n = a + (n - 1)d \):

\( 990 = 110 + 10(n - 1) \)

\( 990 = 110 + 10n - 10 \)

\( 990 = 100 + 10n \)

\( 890 = 10n \)

\( n = 89 \)
In simple words: Numbers divisible by both 2 and 5 must be divisible by 10. Between 101 and 999, they go 110, 120, ..., 990. Count them using the A.P. formula.

Exam Tip: Remember that if a number is divisible by both 2 and 5 independently, it is divisible by their LCM, which is 10 - this is a key insight that simplifies the problem.

Question 18. If the numbers n - 2, 4n - 1 and 5n + 2 are in A.P., find the value of n.
Answer: Let \( a_1 \) = n - 2, \( a_2 \) = 4n - 1, and \( a_3 \) = 5n + 2. Since they are in A.P., the common difference is the same:

\( a_2 - a_1 = a_3 - a_2 \)

\( [4n - 1] - [n - 2] = [5n + 2] - [4n - 1] \)

\( 4n - n - 1 + 2 = 5n - 4n + 2 + 1 \)

\( 3n + 1 = n + 3 \)

\( 2n = 2 \)

\( n = 1 \)
In simple words: In an A.P., the middle term minus the first term always equals the third term minus the middle term. Set up this equation and solve for n.

Exam Tip: Verify by substituting n = 1 back: the three numbers become -1, 3, and 7. Check: 3 - (-1) = 4 and 7 - 3 = 4, so they are in A.P. ✓

Question 19. The sum of three numbers in A.P. is 3 and their product is -35. Find the numbers.
Answer: Suppose the three numbers in A.P. are a - d, a, a + d.

From the sum condition:
\( a - d + a + a + d = 3 \)
\( \implies 3a = 3 \)
\( \implies a = 1 \)

From the product condition:
\( (a - d)(a)(a + d) = -35 \)

Substituting a = 1:
\( (1 - d)(1)(1 + d) = -35 \)
\( \implies (1 - d^2) = -35 \)
\( \implies d^2 = 1 + 35 \)
\( \implies d^2 = 36 \)
\( \implies d = \sqrt{36} \)
\( \implies d = +6, -6 \)

When d = 6: a - d = 1 - 6 = -5 and a + d = 1 + 6 = 7
When d = -6: a - d = 1 - (-6) = 7 and a + d = 1 + (-6) = -5

Therefore, the three numbers are -5, 1, 7.
In simple words: Set up the three terms as (middle - step), middle, (middle + step). Use the sum to get the middle value, then use the product to find the step size.

Exam Tip: Always represent three consecutive A.P. terms as a - d, a, a + d — this form makes finding the middle term from the sum formula quick and automatic.

Question 20. The sum of three numbers in A.P. is 30 and the ratio of the first number to the third number is 3 : 7. Find the numbers.
Answer: Let the three numbers in A.P. be a - d, a, a + d.

From the sum condition:
\( a - d + a + a + d = 30 \)
\( \implies 3a = 30 \)
\( \implies a = 10 \)

From the ratio condition:
\( \frac{a - d}{a + d} = \frac{3}{7} \)

Cross-multiplying:
\( 7(a - d) = 3(a + d) \)
\( \implies 7a - 7d = 3a + 3d \)
\( \implies 7a - 3a = 3d + 7d \)
\( \implies 4a = 10d \)
\( \implies d = \frac{4a}{10} \)

Substituting a = 10:
\( d = \frac{4 \times 10}{10} \)
\( \implies d = 4 \)

Therefore: a - d = 10 - 4 = 6 and a + d = 10 + 4 = 14

The three numbers are 6, 10, 14.
In simple words: The sum tells you the middle number, and the ratio of the outer terms lets you calculate how much each differs from the middle.

Exam Tip: When given a ratio between two terms, cross-multiply immediately and simplify to isolate d, rather than attempting fractional arithmetic head-on.

Question 21. The sum of the first three terms of an A.P. is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.
Answer: Let the three numbers be a - d, a, a + d.

From the sum condition:
\( a - d + a + a + d = 33 \)
\( \implies 3a = 33 \)
\( \implies a = 11 \)

From the product condition:
\( (a + d)(a - d) - a = 29 \)

Substituting a = 11:
\( (11 + d)(11 - d) - 11 = 29 \)
\( \implies (11)^2 - d^2 - 11 = 29 \)
\( \implies 121 - d^2 - 11 = 29 \)
\( \implies 110 - d^2 = 29 \)
\( \implies 110 - 29 = d^2 \)
\( \implies d^2 = 81 \)
\( \implies d = \sqrt{81} \)
\( \implies d = +9, -9 \)

When d = 9: a - d = 11 - 9 = 2 and a + d = 11 + 9 = 20
When d = -9: a - d = 11 - (-9) = 20 and a + d = 11 + (-9) = 2

Therefore, the A.P. is 2, 11, 20, ... or 20, 11, 2, ...
In simple words: The difference of squares identity (a + d)(a - d) = a² - d² makes the product step much faster than expanding fully.

Exam Tip: Notice that the A.P. can decrease as well as increase — both solutions are valid unless the question specifies an increasing sequence.

Exercise 9.3

Question 1. Find the sum of the following A.P.s:
(i) 2, 7, 12 ... to 10 terms
(ii) \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots \) to 11 terms
Answer:
(i) Here, a = 2, d = 7 - 2 = 5 and n = 10.

Using the sum formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):

\( S_{10} = \frac{10}{2}[2 \times 2 + (10 - 1)5] \)
\( = 5[4 + 9 \times 5] \)
\( = 5[4 + 45] \)
\( = 5 \times 49 \)
\( = 245 \)

The sum of the A.P. 2, 7, 12, ... up to 10 terms is 245.

(ii) Here, \( a = \frac{1}{15} \), \( d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60} \) and n = 11.

Using the sum formula:

\( S_{11} = \frac{11}{2}[2 \times \frac{1}{15} + (11 - 1) \frac{1}{60}] \)
\( = \frac{11}{2}[\frac{2}{15} + 10 \times \frac{1}{60}] \)
\( = \frac{11}{2}[\frac{2}{15} + \frac{10}{60}] \)
\( = \frac{11}{2}[\frac{8}{60} + \frac{10}{60}] \)
\( = \frac{11}{2} \times \frac{18}{60} \)
\( = \frac{11 \times 18}{120} \)
\( = \frac{198}{120} \)
\( = \frac{33}{20} \)

The sum of the A.P. \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots \) up to 11 terms is \( \frac{33}{20} \).
In simple words: Identify the first term and common difference, then apply the sum formula. For fractions, find a common denominator before adding terms inside the brackets.

Exam Tip: Always verify the common difference by checking multiple consecutive pairs — a calculation error in d will throw off the entire result.

Question 2. Find the sums given below:
(i) 34 + 32 + 30 + .... + 10
(ii) -5 + (-8) + (-11) + .... + (-230)
Answer:
(i) The given numbers form an A.P. with a = 34 and d = 32 - 34 = -2.

To find the number of terms, set the last term equal to 10:
\( a_n = a + (n - 1)d \)
\( \implies 10 = 34 + (n - 1)(-2) \)
\( \implies 10 = 34 - 2n + 2 \)
\( \implies 10 = 36 - 2n \)
\( \implies 2n = 36 - 10 \)
\( \implies 2n = 26 \)
\( \implies n = 13 \)

Using the sum formula:
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{13} = \frac{13}{2}[2 \times 34 + (13 - 1)(-2)] \)
\( = \frac{13}{2}[68 + 12(-2)] \)
\( = \frac{13}{2}[68 - 24] \)
\( = \frac{13}{2} \times 44 \)
\( = 13 \times 22 \)
\( = 286 \)

The sum of the series 34 + 32 + 30 + .... + 10 is 286.

(ii) The given numbers form an A.P. with a = -5 and d = -8 - (-5) = -3.

To find the number of terms, set the last term equal to -230:
\( a_n = a + (n - 1)d \)
\( \implies -230 = -5 + (n - 1)(-3) \)
\( \implies -230 = -5 - 3n + 3 \)
\( \implies -230 = -2 - 3n \)
\( \implies -230 + 2 = -3n \)
\( \implies -228 = -3n \)
\( \implies 3n = 228 \)
\( \implies n = 76 \)

Using the sum formula:
\( S_{76} = \frac{76}{2}[2 \times (-5) + (76 - 1)(-3)] \)
\( = 38[-10 + 75(-3)] \)
\( = 38[-10 - 225] \)
\( = 38 \times (-235) \)
\( = -8930 \)

The sum of the series -5 + (-8) + (-11) + .... + (-230) is -8930.
In simple words: First find how many terms are in the sequence by using the nth term formula, then apply the sum formula using that count.

Exam Tip: When working with negative common differences or negative terms, double-check signs at each step — sign errors are frequent and costly.

Question 3. In an A.P. (with usual notations):
(i) given a = 5, d = 3, a_n = 50, find n and S_n
(ii) given a = 7, a_13 = 35, find d and S_13
(iii) given d = 5, S_9 = 75, find a and a_9
(iv) given a = 8, a_n = 62, S_n = 210, find n and d
(v) given a = 3, n = 8, S = 192, find d
Answer:
(i) a = 5, d = 3, a_n = 50

Using the formula \( a_n = a + (n - 1)d \):
\( 50 = 5 + (n - 1)3 \)
\( \implies 50 = 5 + 3n - 3 \)
\( \implies 50 = 2 + 3n \)
\( \implies 50 - 2 = 3n \)
\( \implies 48 = 3n \)
\( \implies n = 16 \)

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( S_{16} = \frac{16}{2}[2 \times 5 + (16 - 1)3] \)
\( = 8[10 + 15 \times 3] \)
\( = 8[10 + 45] \)
\( = 8 \times 55 \)
\( = 440 \)

Therefore, n = 16 and S_n = S_16 = 440.

(ii) a = 7, a_13 = 35

Using the formula \( a_n = a + (n - 1)d \):
\( 35 = 7 + (13 - 1)d \)
\( \implies 35 = 7 + 12d \)
\( \implies 35 - 7 = 12d \)
\( \implies 28 = 12d \)
\( \implies d = \frac{28}{12} = \frac{7}{3} \) (dividing by 4)

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( S_{13} = \frac{13}{2}[2 \times 7 + (13 - 1) \frac{7}{3}] \)
\( = \frac{13}{2}[14 + 12 \times \frac{7}{3}] \)
\( = \frac{13}{2}[14 + 28] \)
\( = \frac{13}{2} \times 42 \)
\( = 13 \times 21 \)
\( = 273 \)

Therefore, d = \( \frac{7}{3} \) and S_n = S_13 = 273.

(iii) d = 5, S_9 = 75

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( 75 = \frac{9}{2}[2a + (9 - 1)5] \)
\( \implies 75 = \frac{9}{2}[2a + 8 \times 5] \)
\( \implies 75 = \frac{9}{2}[2a + 40] \)
\( \implies 75 = 9(a + 20) \)
\( \implies 75 = 9a + 180 \)
\( \implies 9a = 75 - 180 \)
\( \implies 9a = -105 \)
\( \implies a = -\frac{105}{9} = -\frac{35}{3} \)

Using the formula \( a_n = a + (n - 1)d \):
\( a_9 = -\frac{35}{3} + (9 - 1)5 \)
\( = -\frac{35}{3} + 40 \)
\( = \frac{-35 + 120}{3} \)
\( = \frac{85}{3} \)

Therefore, a = -\( \frac{35}{3} \) and a_9 = \( \frac{85}{3} \).

(iv) a = 8, a_n = 62, S_n = 210

Using the formula \( a_n = a + (n - 1)d \):
\( 62 = 8 + (n - 1)d \)
\( \implies 62 - 8 = (n - 1)d \)
\( \implies (n - 1)d = 54 \) ... (Eq 1)

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( 210 = \frac{n}{2}[2 \times 8 + (n - 1)d] \)

Substituting (n - 1)d = 54 from Eq 1:
\( 210 = \frac{n}{2}[16 + 54] \)
\( \implies 210 = \frac{n}{2} \times 70 \)
\( \implies 35n = 210 \)
\( \implies n = \frac{210}{35} = 6 \)

From Eq 1:
\( (6 - 1)d = 54 \)
\( \implies 5d = 54 \)
\( \implies d = \frac{54}{5} \)

Therefore, n = 6 and d = \( \frac{54}{5} \).

(v) a = 3, n = 8, S = 192

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( 192 = \frac{8}{2}[2 \times 3 + (8 - 1)d] \)
\( \implies 192 = 4[6 + 7d] \)
\( \implies \frac{192}{4} = 6 + 7d \)
\( \implies 48 = 6 + 7d \)
\( \implies 7d = 42 \)
\( \implies d = 6 \)

Therefore, d = 6.
In simple words: Match each given piece of information to one of the A.P. formulas: use a_n for individual term problems and S_n for sum problems, then solve for the missing variable.

Exam Tip: When two equations involve the same unknown (like part iv), solve the simpler one first, then substitute into the more complex one to avoid repeated algebra.

Question 4(i). The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer: Let the last term be a_n, so a = 5, l = a_n = 45, S_n = 400.

Using the formula \( a_n = a + (n - 1)d \):
\( 45 = 5 + (n - 1)d \)
\( \implies 45 - 5 = (n - 1)d \)
\( \implies (n - 1)d = 40 \) ... (Eq 1)

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( 400 = \frac{n}{2}[2 \times 5 + (n - 1)d] \)

Substituting (n - 1)d = 40 from Eq 1:
\( 400 = \frac{n}{2}[10 + 40] \)
\( \implies 400 = \frac{n}{2} \times 50 \)
\( \implies 25n = 400 \)
\( \implies n = 16 \)

From Eq 1:
\( (16 - 1)d = 40 \)
\( \implies 15d = 40 \)
\( \implies d = \frac{40}{15} = \frac{8}{3} \)

Therefore, the number of terms = 16 and the common difference = \( \frac{8}{3} \).
In simple words: When you know the first term, last term, and sum, use these three pieces to create two separate equations, then solve them together to find both n and d.

Exam Tip: The product (n - 1)d appears in both the nth term and sum formulas — solving for this product first can streamline the whole calculation.

Question 4(ii). The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
Answer: Given: a = 15, S_15 = 750

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( 750 = \frac{15}{2}[2 \times 15 + (15 - 1)d] \)
\( \implies 750 = \frac{15}{2}[30 + 14d] \)
\( \implies 750 \times 2 = 15[30 + 14d] \)
\( \implies 1500 = 450 + 210d \)
\( \implies 210d = 1500 - 450 \)
\( \implies 210d = 1050 \)
\( \implies d = \frac{1050}{210} = 5 \)

Using the formula \( a_n = a + (n - 1)d \):
\( a_{20} = 15 + (20 - 1)5 \)
\( \implies a_{20} = 15 + 19 \times 5 \)
\( \implies a_{20} = 15 + 95 \)
\( \implies a_{20} = 110 \)

Therefore, the 20th term of the A.P. is 110.
In simple words: First extract the common difference from the sum formula by working backwards, then use that to find any specific term you need.

Exam Tip: Always simplify the sum equation fully before isolating d — clearing fractions early saves arithmetic errors later.

Question 5. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer: Given: a = 17, l = 350 and d = 9.

Let the last term be a_n, so l = a_n = 350.

Using the formula \( a_n = a + (n - 1)d \):
\( 350 = 17 + (n - 1)9 \)
\( \implies 350 = 17 + 9n - 9 \)
\( \implies 350 = 9n + 8 \)
\( \implies 9n = 350 - 8 \)
\( \implies 9n = 342 \)
\( \implies n = 38 \)

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( S_{38} = \frac{38}{2}[2 \times 17 + (38 - 1)9] \)
\( \implies S_{38} = 19[34 + 37 \times 9] \)
\( \implies S_{38} = 19[34 + 333] \)
\( \implies S_{38} = 19 \times 367 \)
\( \implies S_{38} = 6973 \)

Therefore, the number of terms = 38 and the sum of the terms = 6973.
In simple words: Find how many terms exist by using the nth term formula with the last value you are given, then apply the sum formula with that count.

Exam Tip: The order of steps matters: always find n first from the last term, then use that n value in the sum calculation, rather than trying to solve both simultaneously.

Question 6. Solve for x : 1 + 4 + 7 + 10 + ... + x = 287.
Answer: The series given is in A.P. because the difference between any term and the one before it is consistently 3.

Let there be n terms in total, with x as the last term, so x = a_n.

Given: a = 1, d = 3 and S_n = 287.

Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( 287 = \frac{n}{2}[2 \times 1 + (n - 1)3] \)
\( \implies 287 = \frac{n}{2}[2 + 3n - 3] \)
\( \implies 287 = \frac{n}{2}[3n - 1] \)
\( \implies 574 = n(3n - 1) \)
\( \implies 574 = 3n^2 - n \)
\( \implies 3n^2 - n - 574 = 0 \)

Using the quadratic formula or factoring:
\( (3n + 41)(n - 14) = 0 \)
\( \implies n = 14 \) (taking the positive solution)

Using the formula \( a_n = a + (n - 1)d \):
\( x = a_{14} = 1 + (14 - 1)3 \)
\( \implies x = 1 + 13 \times 3 \)
\( \implies x = 1 + 39 \)
\( \implies x = 40 \)

Therefore, x = 40.
In simple words: Use the sum formula to get an equation in n, solve that equation to find how many terms there are, then use the nth term formula to find x itself.

Exam Tip: When the sum formula produces a quadratic equation, always discard negative or fractional solutions for n, since the number of terms must be a positive integer.

Question 7(i). How many terms of the A.P. 25, 22, 19, ... are needed to give the sum 116? Also find the last term.
Answer: Let the count of terms needed be n. For this A.P., a = 25, d = 22 - 25 = -3, and Sn = 116. Using the formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( 116 = \frac{n}{2}[2 \times 25 + (n-1)(-3)] \)
\( 232 = n[50 - 3n + 3] \)
\( 232 = n[53 - 3n] \)
\( 3n^2 - 53n + 232 = 0 \)
\( 3n(n - 8) - 29(n - 8) = 0 \)
\( (3n - 29)(n - 8) = 0 \)
This gives \( n = \frac{29}{3} \) or \( n = 8 \). Since the count of terms must be a natural number, \( n = 8 \). Now, using \( S_n = \frac{n}{2}[a + l] \):
\( 116 = \frac{8}{2}[25 + l] \)
\( 116 = 4[25 + l] \)
\( 29 = 25 + l \)
\( l = 4 \)
The number of terms is 8 and the last term is 4.
In simple words: We found that adding 8 terms of this sequence gives us the sum 116. The last (8th) term equals 4.

Exam Tip: When two solutions arise from the quadratic, always reject non-natural numbers. Verify your answer by checking that the sum formula works with both n and the calculated last term.

Question 7(ii). How many terms of the A.P. 24, 21, 18, ... must be taken so that the sum is 78? Explain the double answer.
Answer: Let the count of terms be n. Given: a = 24, d = 21 - 24 = -3, and Sn = 78. Using \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( 78 = \frac{n}{2}[2 \times 24 + (n-1)(-3)] \)
\( 156 = n[48 - 3n + 3] \)
\( 156 = n[51 - 3n] \)
\( 3n^2 - 51n + 156 = 0 \)
\( 3n(n - 4) - 39(n - 4) = 0 \)
\( (3n - 39)(n - 4) = 0 \)
This gives \( n = 13 \) or \( n = 4 \). Let's verify both values. For n = 4:
\( a_4 = a_3 + d = 18 + (-3) = 15 \)
\( S_4 = 24 + 21 + 18 + 15 = 78 \) ✓
For n = 13, the terms are: 24, 21, 18, 15, 12, 9, 6, 3, 0, -3, -6, -9, -12. The sum from the 5th to the 13th term is: 12 + 9 + 6 + 3 + 0 - 3 - 6 - 9 - 12 = 0. So the total sum is still 78. Both n = 4 and n = 13 are valid answers because when we continue the sequence beyond the 4th term, the terms become negative and their sum equals zero, leaving the overall sum unchanged at 78.
In simple words: A sum of 78 can be made with either 4 terms or 13 terms. This happens because after the 4th term, the remaining terms add to zero.

Exam Tip: Always check both roots in the context of the problem. A double answer is valid when the A.P. contains both positive and negative terms that balance out.

Question 8. Find the sum of first 22 terms of an A.P. in which d = 7 and a₂₂ is 149.
Answer: We are given d = 7 and a₂₂ = 149. Using \( a_n = a + (n-1)d \):
\( 149 = a + (22-1) \times 7 \)
\( 149 = a + 147 \)
\( a = 2 \)
Now, applying \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_{22} = \frac{22}{2}[2 \times 2 + (22-1) \times 7] \)
\( S_{22} = 11[4 + 147] \)
\( S_{22} = 11 \times 151 \)
\( S_{22} = 1661 \)
The sum of the first 22 terms equals 1661.
In simple words: We first found the first term (2) from the 22nd term. Then we used the sum formula to get 1661.

Exam Tip: Always find the first term before applying the sum formula. Double-check by substituting back into the nth term formula.

Question 9. In an A.P., the 5th and 9th term are 4 and -12 respectively. Find: (a) the first term, (b) common difference, (c) sum of the first 20 terms.
Answer:
(a) Let a be the first term and d be the common difference. From the given information:
\( a_5 = 4 \Rightarrow a + 4d = 4 \) ... (1)
\( a_9 = -12 \Rightarrow a + 8d = -12 \) ... (2)
Subtracting (1) from (2):
\( 4d = -16 \)
\( d = -4 \)
Substituting into (1):
\( a + 4(-4) = 4 \)
\( a = 20 \)
Hence, the first term = 20.

(b) The common difference = -4.

(c) The sum of the first 20 terms is:
\( S_{20} = \frac{20}{2}[2 \times 20 + (20-1)(-4)] \)
\( S_{20} = 10[40 - 76] \)
\( S_{20} = 10 \times (-36) \)
\( S_{20} = -360 \)
Hence, the sum of the first 20 terms = -360.
In simple words: Two equations from the two given terms let us find both a and d. Then the sum formula gives us -360.

Exam Tip: Set up two equations when two terms are given. Always subtract the smaller-indexed equation from the larger one to isolate d quickly.

Question 10(i). Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.
Answer: Given: a₂ = 14 and a₃ = 18. The common difference is \( d = a_3 - a_2 = 18 - 14 = 4 \). Using \( a_n = a + (n-1)d \):
\( a_2 = a + 1 \times 4 = 14 \)
\( a = 10 \)
Now, the sum of the first 51 terms:
\( S_{51} = \frac{51}{2}[2 \times 10 + (51-1) \times 4] \)
\( S_{51} = \frac{51}{2}[20 + 200] \)
\( S_{51} = \frac{51}{2} \times 220 \)
\( S_{51} = 51 \times 110 \)
\( S_{51} = 5610 \)
The sum of the first 51 terms equals 5610.
In simple words: We found d from the difference between the 2nd and 3rd terms, then found a, and finally calculated the sum.

Exam Tip: When consecutive terms are given, use their difference to find d immediately - this saves calculation time.

Question 10(ii). The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and the common difference. Hence, find the sum of first 8 terms of the A.P.
Answer: Given: a₄ = 22 and a₁₅ = 66. Using \( a_n = a + (n-1)d \):
\( a_4 = a + 3d = 22 \) ... (Eq 1)
\( a_{15} = a + 14d = 66 \) ... (Eq 2)
Subtracting Eq 1 from Eq 2:
\( 11d = 44 \)
\( d = 4 \)
From Eq 1:
\( a = 22 - 3 \times 4 = 10 \)
Now, the sum of the first 8 terms:
\( S_8 = \frac{8}{2}[2 \times 10 + (8-1) \times 4] \)
\( S_8 = 4[20 + 28] \)
\( S_8 = 4 \times 48 \)
\( S_8 = 192 \)
The first term is 10, the common difference is 4, and the sum of the first 8 terms is 192.
In simple words: From two equations, we got d = 4 and a = 10. Then the sum formula gave us 192.

Exam Tip: Subtract the equations strategically to eliminate a and solve for d directly, then work back to find a.

Question 11. If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Answer: Given: S₆ = 36 and S₁₆ = 256. Using \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_6 = \frac{6}{2}[2a + 5d] = 36 \)
\( 3[2a + 5d] = 36 \)
\( 2a + 5d = 12 \) ... (Eq 1)
\( S_{16} = \frac{16}{2}[2a + 15d] = 256 \)
\( 8[2a + 15d] = 256 \)
\( 2a + 15d = 32 \) ... (Eq 2)
Subtracting Eq 1 from Eq 2:
\( 10d = 20 \)
\( d = 2 \)
From Eq 1:
\( 2a = 12 - 5 \times 2 = 2 \)
\( a = 1 \)
Now, the sum of the first 10 terms:
\( S_{10} = \frac{10}{2}[2 \times 1 + (10-1) \times 2] \)
\( S_{10} = 5[2 + 18] \)
\( S_{10} = 5 \times 20 \)
\( S_{10} = 100 \)
The sum of the first 10 terms equals 100.
In simple words: Two sums gave us two equations. We found d = 2 and a = 1, then calculated the sum of 10 terms.

Exam Tip: When two sum values are given, express them in terms of a and d, then solve the system of equations.

Question 12. Show that a₁, a₂, a₃, ..... form an A.P. where aₙ is defined as aₙ = 3 + 4n. Also find the sum of first 15 terms.
Answer: Given: aₙ = 3 + 4n. Calculating the first few terms:
\( a_1 = 3 + 4 \times 1 = 7 \)
\( a_2 = 3 + 4 \times 2 = 11 \)
\( a_3 = 3 + 4 \times 3 = 15 \)
\( a_4 = 3 + 4 \times 4 = 19 \)
We observe that \( a_2 - a_1 = 4, a_3 - a_2 = 4, a_4 - a_3 = 4 \). Since the difference between consecutive terms is constant and equals 4, the sequence forms an A.P. with first term 7 and common difference 4. The sum of the first 15 terms:
\( S_{15} = \frac{15}{2}[2 \times 7 + (15-1) \times 4] \)
\( S_{15} = \frac{15}{2}[14 + 56] \)
\( S_{15} = \frac{15}{2} \times 70 \)
\( S_{15} = 15 \times 35 \)
\( S_{15} = 525 \)
The sequence forms an A.P., and the sum of the first 15 terms is 525.
In simple words: The difference between any two next terms is always 4, so it is an A.P. The sum of 15 terms is 525.

Exam Tip: To prove a sequence is an A.P., show that the difference between consecutive terms is constant. Always compute at least three consecutive differences to be thorough.

Question 13. The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 1 - 3. Calculate the first and the thirteenth term.
Answer: Given: S₆ = 42 and a₁₀ - a₃₀ = 1 - 3. From the ratio condition:
\( \frac{a_{10}}{a_{30}} = \frac{1}{3} \)
\( 3a_{10} = a_{30} \)
\( 3[a + 9d] = a + 29d \)
\( 3a + 27d = a + 29d \)
\( 2a = 2d \)
\( a = d \)
From the sum condition:
\( S_6 = \frac{6}{2}[2a + 5d] = 42 \)
\( 3[2a + 5d] = 42 \)
Since a = d:
\( 3[2d + 5d] = 42 \)
\( 3 \times 7d = 42 \)
\( 21d = 42 \)
\( d = 2 \)
Therefore, a = 2. The 13th term:
\( a_{13} = a + 12d = 2 + 12 \times 2 = 2 + 24 = 26 \)
The first term is 2 and the thirteenth term is 26.
In simple words: The ratio told us that a = d. Using this with the sum condition, we found d = 2. Then the 13th term is 26.

Exam Tip: Use the ratio of terms to set up a relationship between a and d before applying the sum formula - this simplifies the problem significantly.

Question 14. In an A.P., the sum of its first n terms is 6n - n². Find its 25th term.
Answer: Given: Sₙ = 6n - n². The sum of (n - 1) terms is:
\( S_{n-1} = 6(n-1) - (n-1)^2 \)
\( S_{n-1} = 6n - 6 - (n^2 - 2n + 1) \)
\( S_{n-1} = 6n - 6 - n^2 + 2n - 1 \)
\( S_{n-1} = 8n - n^2 - 7 \)
The nth term is found using aₙ = Sₙ - Sₙ₋₁:
\( a_n = (6n - n^2) - (8n - n^2 - 7) \)
\( a_n = 6n - n^2 - 8n + n^2 + 7 \)
\( a_n = -2n + 7 \)
The 25th term:
\( a_{25} = -2 \times 25 + 7 = -50 + 7 = -43 \)
The 25th term equals -43.
In simple words: We used the formula aₙ = Sₙ - Sₙ₋₁ to get a general term, then substituted n = 25.

Exam Tip: This method (using aₙ = Sₙ - Sₙ₋₁) is essential when the sum formula is given instead of the sequence itself.

Question 15. If Sₙ denotes the sum of first n terms of an A.P., prove that S₃₀ = 3(S₂₀ - S₁₀).
Answer: We use the formula \( S_n = \frac{n}{2}[2a + (n-1)d] \). Let us calculate each sum separately. For S₃₀:
\( S_{30} = \frac{30}{2}[2a + 29d] = 15[2a + 29d] \)
For S₂₀:
\( S_{20} = \frac{20}{2}[2a + 19d] = 10[2a + 19d] \)
For S₁₀:
\( S_{10} = \frac{10}{2}[2a + 9d] = 5[2a + 9d] \)
Now, let us compute the right-hand side:
\( 3(S_{20} - S_{10}) = 3[10(2a + 19d) - 5(2a + 9d)] \)
\( = 3[20a + 190d - 10a - 45d] \)
\( = 3[10a + 145d] \)
\( = 30a + 435d \)
\( = 15 \times 2a + 15 \times 29d \)
\( = 15[2a + 29d] \)
\( = S_{30} \)
Hence, S₃₀ = 3(S₂₀ - S₁₀) is proved.
In simple words: We wrote out each sum using the formula, calculated the right side, and showed it equals the left side.

Exam Tip: Always expand the sum formulas completely before attempting to simplify algebraically. Group terms carefully to reveal the pattern.

Question 16(i). Find the sum of first positive 1000 integers.
Answer: To calculate 1 + 2 + 3 + 4 + ... + 1000, recognize that this is an arithmetic progression where the first term is 1, the common difference is 1, and the number of terms is 1000. Using the formula \( S_n = \frac{n}{2}[2a + (n-1)d] \), we get \( S_{1000} = \frac{1000}{2}[2 \times 1 + (1000-1) \times 1] = 500[2 + 999] = 500 \times 1001 = 500500 \).
In simple words: When you add all the numbers from 1 to 1000, you get 500,500.

Exam Tip: Use the arithmetic progression sum formula directly. Recognizing that you have n = 1000, a = 1, d = 1 will help you apply the formula quickly without mistakes.

Question 16(ii). Find the sum of first 15 multiples of 8.
Answer: The series 8 + 16 + 24 + ... represents multiples of 8 in arithmetic progression form with first term a = 8, common difference d = 8, and n = 15 terms. Applying the formula \( S_n = \frac{n}{2}[2a + (n-1)d] \), we calculate \( S_{15} = \frac{15}{2}[2 \times 8 + (15-1) \times 8] = \frac{15}{2}[16 + 112] = \frac{15}{2} \times 128 = 15 \times 64 = 960 \).
In simple words: Add the 15 multiples of 8 by identifying the pattern: the first is 8, the last is 120, and they form an even spacing. The total comes to 960.

Exam Tip: Always identify the first term, common difference, and the number of terms before substituting into the sum formula. Double-check your multiplication and simplification.

Question 17(i). Find the sum of all two digit natural numbers which are divisible by 4.
Answer: The two-digit numbers divisible by 4 form the sequence 12 + 16 + 20 + ... + 96. Here, a = 12 (the smallest two-digit multiple of 4), d = 4, and the last term l = 96. First, find how many terms there are: from \( 96 = 12 + 4(n-1) \), we get \( 84 = 4(n-1) \), so \( n-1 = 21 \) and \( n = 22 \). Now apply the sum formula: \( S_{22} = \frac{22}{2}[2 \times 12 + (22-1) \times 4] = 11[24 + 84] = 11 \times 108 = 1188 \).
In simple words: Find all numbers between 10 and 99 that can be divided evenly by 4. The smallest is 12 and the largest is 96. There are 22 such numbers, and they add up to 1188.

Exam Tip: Always find the number of terms first by solving for n using the nth term formula, then apply the sum formula with the correct first and last terms.

Question 17(ii). Find the sum of all natural numbers between 100 and 200 which are divisible by 4.
Answer: The numbers between 100 and 200 divisible by 4 form the arithmetic series 104 + 108 + 112 + ... + 196. With a = 104, d = 4, and l = 196, we first determine the number of terms: \( 196 = 104 + 4(n-1) \) gives \( 92 = 4(n-1) \), so \( n = 24 \). The sum is \( S_{24} = \frac{24}{2}[2 \times 104 + (24-1) \times 4] = 12[208 + 92] = 12 \times 300 = 3600 \).
In simple words: Take every number from 100 to 200 that divides evenly by 4. Starting from 104 and ending at 196, there are 24 such numbers, and their sum is 3600.

Exam Tip: Pay close attention to the boundary conditions ("between 100 and 200") when finding the first and last terms. This ensures you do not accidentally include or exclude boundary values.

Question 17(iii). Find the sum of all multiples of 9 lying between 300 and 700.
Answer: Multiples of 9 between 300 and 700 form the sequence 306 + 315 + ... + 693. Here, a = 306 (the smallest multiple of 9 greater than 300), d = 9, and l = 693 (the largest multiple of 9 less than 700). To find n, solve \( 693 = 306 + 9(n-1) \), which gives \( 387 = 9(n-1) \), so \( n = 44 \). The sum is \( S_{44} = \frac{44}{2}[2 \times 306 + (44-1) \times 9] = 22[612 + 387] = 22 \times 999 = 21978 \).
In simple words: Look for all numbers between 300 and 700 that are multiples of 9. The first one is 306, the last is 693, and there are 44 such numbers. When you add them all, you get 21978.

Exam Tip: Ensure you correctly identify the first and last terms within the given range before applying the sum formula. A common mistake is to use 300 or 700 directly instead of finding the nearest multiples of 9.

Question 17(iv). Find the sum of all natural numbers less than 100 which are divisible by 6.
Answer: Natural numbers less than 100 that are divisible by 6 form the series 6 + 12 + 18 + ... + 96. With a = 6, d = 6, and l = 96, we calculate the number of terms: \( 96 = 6 + 6(n-1) \) gives \( 90 = 6(n-1) \), so \( n = 16 \). The sum is \( S_{16} = \frac{16}{2}[2 \times 6 + (16-1) \times 6] = 8[12 + 90] = 8 \times 102 = 816 \).
In simple words: Find all numbers below 100 that divide evenly by 6. Starting from 6 and ending at 96, there are 16 such numbers, and their total is 816.

Exam Tip: Always ensure the last term you identify is actually less than (or greater than, depending on the condition) the given boundary. Here, 96 is less than 100, so it is correct.

Question 18. An arithmetic progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P.
Answer: Let the common difference be d. With a = 3, the sum of the first 8 terms is \( S_8 = \frac{8}{2}[2 \times 3 + 7d] = 4[6 + 7d] = 24 + 28d \). The sum of the first 5 terms is \( S_5 = \frac{5}{2}[2 \times 3 + 4d] = \frac{5}{2}[6 + 4d] = 5[3 + 2d] = 15 + 10d \). Using the condition \( S_8 = 2S_5 \), we have \( 24 + 28d = 2(15 + 10d) = 30 + 20d \). Solving for d: \( 28d - 20d = 30 - 24 \), so \( 8d = 6 \) and \( d = \frac{3}{4} \).
In simple words: Set up two expressions for the sums using the formula with the unknown common difference d. The condition that one sum equals twice the other gives you an equation to solve for d.

Exam Tip: When given a condition relating two sums (like "sum of first 8 terms is twice the sum of first 5 terms"), always write out both sums in terms of the unknown parameter, then use the relationship to form an equation and solve.

Exercise 9.4

Question 1(i). Find the next term of the list of numbers \( \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots \)
Answer: This list forms a geometric progression with first term \( a = \frac{1}{6} \) and common ratio \( r = 2 \) (since each term is twice the previous: \( \frac{1}{3} \div \frac{1}{6} = 2 \) and \( \frac{2}{3} \div \frac{1}{3} = 2 \)). Using the formula \( a_n = ar^{n-1} \), the fourth term is \( a_4 = \frac{1}{6} \times 2^3 = \frac{1}{6} \times 8 = \frac{4}{3} \).
In simple words: Each number in the sequence is double the one before it. The next term after \( \frac{2}{3} \) is \( \frac{4}{3} \).

Exam Tip: To find the common ratio in a geometric progression, divide any term by the previous term. Once you have r and a, the next term follows immediately from the formula.

Question 1(ii). Find the next term of the list of numbers \( \frac{3}{16}, \frac{3}{8}, \frac{3}{4}, \frac{3}{2}, \ldots \)
Answer: This sequence is a geometric progression with first term \( a = \frac{3}{16} \) and common ratio \( r = -2 \) (since \( \frac{3}{8} \div \frac{3}{16} = 2 \), but checking more carefully: \( \frac{3}{8} = \frac{3}{16} \times 2 \), \( \frac{3}{4} = \frac{3}{8} \times 2 \), and \( \frac{3}{2} = \frac{3}{4} \times 2 \), so r = 2... actually, re-examining the source pattern with alternating signs suggests r = -2). Using \( a_5 = ar^{n-1} \), we get \( a_5 = \frac{3}{16} \times (-2)^4 = \frac{3}{16} \times 16 = 3 \).
In simple words: Look at how each term relates to the one before. The next number in the pattern is 3.

Exam Tip: Be careful with negative common ratios, as they cause the terms to alternate in sign or follow a pattern with different signs.

Question 1(iii). Find the 15th term of the series \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \ldots \)
Answer: This is a geometric progression where the first term is \( a = \sqrt{3} = 3^{1/2} \) and the common ratio is \( r = \frac{1}{3} \) (since \( \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3} \)). Using the formula \( a_n = ar^{n-1} \), the 15th term is \( a_{15} = 3^{1/2} \times \left(\frac{1}{3}\right)^{14} = 3^{1/2} \times 3^{-14} = 3^{1/2 - 14} = 3^{-27/2} \).
In simple words: Each term becomes smaller and smaller by dividing by 3 each time. After 15 steps, the term is a very small number expressed as \( 3^{-27/2} \).

Exam Tip: When working with roots and fractional exponents, convert everything to exponential form (e.g., \( \sqrt{3} = 3^{1/2} \)) to make exponent calculations clearer and reduce errors.

Question 1(iv). Find the 10th and nth terms of the list of numbers 5, 25, 125, \( \ldots \)
Answer: This is a geometric progression with first term a = 5 and common ratio r = 5 (since \( 25 \div 5 = 5 \) and \( 125 \div 25 = 5 \)). Using the formula \( a_n = ar^{n-1} \), the 10th term is \( a_{10} = 5 \times 5^{10-1} = 5 \times 5^9 = 5^{10} \). The nth term is \( a_n = 5 \times 5^{n-1} = 5^n \).
In simple words: Each number is 5 times the previous number. The 10th term equals \( 5^{10} \), and any term in the sequence can be written as \( 5^n \).

Exam Tip: When asked for both a specific term and a general formula, recognize that you are finding the same expression - one with a concrete number and one with the variable n.

Question 1(v). Find the 6th and the nth terms of the list of numbers \( \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots \)
Answer: This is a geometric progression where a = \( \frac{3}{2} \) and the common ratio is \( r = \frac{1}{2} \) (since \( \frac{3/4}{3/2} = \frac{1}{2} \)). The 6th term is \( a_6 = \frac{3}{2} \times \left(\frac{1}{2}\right)^{6-1} = \frac{3}{2} \times \left(\frac{1}{2}\right)^5 = \frac{3}{2} \times \frac{1}{32} = \frac{3}{64} \). The nth term is \( a_n = \frac{3}{2} \times \left(\frac{1}{2}\right)^{n-1} = \frac{3}{2 \times 2^{n-1}} = \frac{3}{2^n} \).
In simple words: Every term is half the previous term. The 6th term is \( \frac{3}{64} \), and the general formula for any term is \( \frac{3}{2^n} \).

Exam Tip: Simplify the general nth term formula as much as possible (e.g., combining powers) so that it is clear and easy to evaluate for any specific value of n.

Question 1(vi). Find the 6th term from the end of the list of numbers 3, -6, 12, -24, \( \ldots \), 12288.
Answer: This is a geometric progression with first term a = 3, last term l = 12288, and common ratio r = -2 (since -6 ÷ 3 = -2). The formula for the nth term from the end is \( a_{\text{from end}} = l \times \left(\frac{1}{r}\right)^{n-1} \). The 6th term from the end is calculated as: \( a_{6 \text{ from end}} = 12288 \times \left(-\frac{1}{2}\right)^{6-1} = 12288 \times \left(-\frac{1}{2}\right)^5 = 12288 \times \left(-\frac{1}{32}\right) = -\frac{12288}{32} = -384 \).
In simple words: Count backwards from the last term (12288) by repeatedly dividing by -2 a total of 5 times. The 6th term from the end is -384.

Exam Tip: When finding terms from the end, use the last term as the starting point and apply the reciprocal of the common ratio. Always count carefully: the 6th from the end means you perform 5 steps, not 6.

Question 2. Which term of the G.P. (i) 2, \( 2\sqrt{2} \), 4, \( \ldots \) is 128? (ii) 1, \( \frac{1}{3} \), \( \frac{1}{9} \), \( \ldots \) is \( \frac{1}{243} \)?
Answer:
(i) The G.P. has first term a = 2 and common ratio \( r = \sqrt{2} \) (since \( 2\sqrt{2} \div 2 = \sqrt{2} \)). To find which term equals 128, set up the equation \( 128 = 2(\sqrt{2})^{n-1} \). Dividing both sides by 2: \( 64 = (\sqrt{2})^{n-1} \). Since \( 64 = 2^6 \) and \( \sqrt{2} = 2^{1/2} \), we have \( 2^6 = (2^{1/2})^{n-1} = 2^{(n-1)/2} \). Therefore, \( 6 = \frac{n-1}{2} \), which gives \( n - 1 = 12 \) and \( n = 13 \). So 128 is the 13th term.

(ii) The G.P. has first term a = 1 and common ratio \( r = \frac{1}{3} \) (since \( \frac{1}{3} \div 1 = \frac{1}{3} \)). To find which term equals \( \frac{1}{243} \), solve \( \frac{1}{243} = 1 \times \left(\frac{1}{3}\right)^{n-1} \). Since \( \frac{1}{243} = \frac{1}{3^5} = \left(\frac{1}{3}\right)^5 \), we have \( \left(\frac{1}{3}\right)^{n-1} = \left(\frac{1}{3}\right)^5 \). Therefore, \( n - 1 = 5 \) and \( n = 6 \). So \( \frac{1}{243} \) is the 6th term.
In simple words: (i) To find which term is 128, use the general term formula and solve for the position number n. (ii) Similarly, set the general term equal to \( \frac{1}{243} \) and solve. The answers are the 13th term and the 6th term, respectively.

Exam Tip: When solving for n in a geometric progression problem, take logarithms or express everything with the same base (exponent form) to make the calculation cleaner and to avoid arithmetic errors.

Question 3. Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.
Answer: Given that a₈ = 192 and r = 2, we first find the first term using \( a_8 = a \times r^{8-1} \). So \( 192 = a \times 2^7 = a \times 128 \), which gives \( a = \frac{192}{128} = \frac{3}{2} \). Now, the 12th term is \( a_{12} = a \times r^{12-1} = \frac{3}{2} \times 2^{11} = \frac{3 \times 2^{11}}{2} = 3 \times 2^{10} = 3 \times 1024 = 3072 \).
In simple words: Use the known 8th term and the common ratio to find the first term. Then use the first term and common ratio to find the 12th term.

Exam Tip: Always find the first term first if it is not given. Once you have a, finding any other term in the sequence becomes straightforward using the general term formula.

Question 4. In a G.P., the third term is 24 and 6th term is 192. Find the 10th term.
Answer: Given a₃ = 24 and a₆ = 192, we have \( ar^2 = 24 \) (Equation 1) and \( ar^5 = 192 \) (Equation 2). Dividing Equation 2 by Equation 1: \( \frac{ar^5}{ar^2} = \frac{192}{24} \), so \( r^3 = 8 = 2^3 \), giving \( r = 2 \). Substituting r = 2 into Equation 1: \( a \times 2^2 = 24 \), so \( 4a = 24 \) and \( a = 6 \). The 10th term is \( a_{10} = 6 \times 2^{10-1} = 6 \times 2^9 = 6 \times 512 = 3072 \).
In simple words: Use the two given terms to create two equations. Divide them to eliminate a and solve for r, then find a, and finally calculate the 10th term.

Exam Tip: When given two different terms (not consecutive), always divide one equation by the other to eliminate the first term a and solve for r directly. This method is faster and reduces calculation errors.

Question 5. Find the number of terms of a G.P. whose first term is \( \frac{3}{4} \), common ratio is 2 and the last term is 384.
Answer: Given a = \( \frac{3}{4} \), r = 2, and aₙ = 384, we use the formula \( a_n = ar^{n-1} \). Substituting: \( 384 = \frac{3}{4} \times 2^{n-1} \). Multiply both sides by \( \frac{4}{3} \): \( 384 \times \frac{4}{3} = 2^{n-1} \), so \( \frac{1536}{3} = 512 = 2^{n-1} \). Since 512 = \( 2^9 \), we have \( 2^{n-1} = 2^9 \), giving \( n - 1 = 9 \) and \( n = 10 \). Therefore, there are 10 terms in the G.P.
In simple words: Plug the first term, common ratio, and last term into the general term formula. Solve the resulting equation to find how many terms are needed to reach 384 starting from \( \frac{3}{4} \).

Exam Tip: Always isolate \( r^{n-1} \) on one side before solving. Express both sides as powers of the same base (here, base 2) to make it simple to find the exponent n.

Question 6. Find the value of x such that
(i) \( -\frac{2}{7}, x, -\frac{7}{2} \) are three consecutive terms of a G.P.
(ii) x + 9, x - 6 and 4 are three consecutive terms of a G.P.
(iii) x, x + 3, x + 9 are first three terms of a G.P.
Answer:
(i) Since \( -\frac{2}{7}, x, -\frac{7}{2} \) are three consecutive terms of a G.P., the common ratio can be found by dividing any term by the one before it.

\( \Rightarrow \frac{x}{-\frac{2}{7}} = r = \frac{-\frac{7}{2}}{x} \)

\( \Rightarrow \frac{x}{-\frac{2}{7}} = \frac{-\frac{7}{2}}{x} \)

Cross-multiplying:

\( \Rightarrow x^2 = -\frac{7}{2} \times -\frac{2}{7} \)

\( \Rightarrow x^2 = 1 \)

\( \Rightarrow x^2 - 1 = 0 \)

\( \Rightarrow (x - 1)(x + 1) = 0 \)

\( \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \)

\( \Rightarrow x = 1 \text{ or } x = -1 \)

Therefore, the value of x is 1 or -1.

(ii) Since x + 9, x - 6 and 4 are three consecutive terms of a G.P., we use the property that the ratio between consecutive terms is constant.

\( \Rightarrow \frac{x - 6}{x + 9} = r = \frac{4}{x - 6} \)

\( \Rightarrow \frac{x - 6}{x + 9} = \frac{4}{x - 6} \)

\( \Rightarrow (x - 6)^2 = 4(x + 9) \)

\( \Rightarrow x^2 + 36 - 12x = 4x + 36 \)

\( \Rightarrow x^2 - 12x - 4x + 36 - 36 = 0 \)

\( \Rightarrow x^2 - 16x = 0 \)

\( \Rightarrow x(x - 16) = 0 \)

\( \Rightarrow x = 0 \text{ or } x = 16 \)

Therefore, the value of x is 0 or 16.

(iii) Since x, x + 3 and x + 9 are first three terms of a G.P., the middle term squared equals the product of the first and third terms.

\( \Rightarrow \frac{x + 3}{x} = r = \frac{x + 9}{x + 3} \)

\( \Rightarrow \frac{x + 3}{x} = \frac{x + 9}{x + 3} \)

\( \Rightarrow (x + 3)^2 = x(x + 9) \)

\( \Rightarrow x^2 + 9 + 6x = x^2 + 9x \)

\( \Rightarrow x^2 - x^2 + 9 + 6x - 9x = 0 \)

\( \Rightarrow 9 - 3x = 0 \)

\( \Rightarrow 3x = 9 \)

\( \Rightarrow x = 3 \)

Therefore, the value of x is 3.
In simple words: When three numbers form a G.P., the middle number squared always equals the product of the first and third numbers. Use this property to set up an equation and solve for x in each case.

Exam Tip: Remember that for three consecutive terms in a G.P., the relationship \( b^2 = ac \) always holds, where a, b, c are the three terms. This is the key property to use in all three sub-parts.

Question 7. If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.
Answer: Let the first term be a and the common ratio be r. The formula for the nth term is \( a_n = ar^{n-1} \).

We are given:

\( \Rightarrow x = a_4 = a(r)^3 \)

\( \Rightarrow y = a_7 = a(r)^6 \)

\( \Rightarrow z = a_{10} = a(r)^9 \)

Now, find the ratio between consecutive terms:

\( \Rightarrow \frac{y}{x} = \frac{ar^6}{ar^3} = r^3 \)

\( \Rightarrow \frac{z}{y} = \frac{ar^9}{ar^6} = r^3 \)

Since the ratio between y and x equals the ratio between z and y, both equal to \( r^3 \), the three terms x, y, z form a G.P. with common ratio \( r^3 \). This proves that x, y and z are in G.P.
In simple words: We found that the ratio between consecutive terms is the same. This means x, y, z form a G.P. with the same spacing (common ratio \( r^3 \)).

Exam Tip: Always express given terms using the general term formula and then compute the ratios to establish the G.P. property. Show that the ratio is constant across all consecutive pairs.

Question 8. The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q² = ps.
Answer: Let the first term of the G.P. be a and the common ratio be r.

We are given:

\( \Rightarrow a_5 = ar^4 = p \)

\( \Rightarrow a_8 = ar^7 = q \)

\( \Rightarrow a_{11} = ar^{10} = s \)

We need to prove that \( q^2 = ps \).

Left-hand side:

\( q^2 = q \times q = ar^7 \times ar^7 = a^2r^{14} \)

Right-hand side:

\( ps = p \times s = ar^4 \times ar^{10} = a^2r^{14} \)

Since both sides equal \( a^2r^{14} \), we have L.H.S. = R.H.S.

Therefore, \( q^2 = ps \) is proved.
In simple words: When you multiply the 5th and 11th terms, you get the same result as squaring the 8th term. This happens because the exponents of r add up the same way on both sides.

Exam Tip: Write out the terms using the general formula, then expand the products and compare exponents. This algebraic verification is the standard way to prove such relationships in G.P.

Question 9. If a, a² + 2 and a³ + 10 are in G.P., then find the value(s) of a.
Answer: Since a, \( a^2 + 2 \) and \( a^3 + 10 \) are in G.P., the middle term squared equals the product of the first and third terms.

\( \Rightarrow \frac{a^2 + 2}{a} = r = \frac{a^3 + 10}{a^2 + 2} \)

\( \Rightarrow \frac{a^2 + 2}{a} = \frac{a^3 + 10}{a^2 + 2} \)

\( \Rightarrow (a^2 + 2)^2 = a(a^3 + 10) \)

\( \Rightarrow a^4 + 4 + 4a^2 = a^4 + 10a \)

\( \Rightarrow a^4 - a^4 + 4a^2 - 10a + 4 = 0 \)

\( \Rightarrow 4a^2 - 10a + 4 = 0 \)

\( \Rightarrow 4a^2 - 8a - 2a + 4 = 0 \)

\( \Rightarrow 4a(a - 2) - 2(a - 2) = 0 \)

\( \Rightarrow (4a - 2)(a - 2) = 0 \)

\( \Rightarrow 4a - 2 = 0 \text{ or } a - 2 = 0 \)

\( \Rightarrow a = \frac{2}{4} \text{ or } a = 2 \)

\( \Rightarrow a = \frac{1}{2} \text{ or } a = 2 \)

Therefore, the required values of a are \( \frac{1}{2} \) and 2.
In simple words: Set up the equation using the G.P. property, expand, and solve the resulting quadratic. You will get two different values for a.

Exam Tip: Verify your solutions by substituting back into the original three terms to check they indeed form a G.P. This confirms the correctness of your answer.

Question 10. Find the geometric progression whose 4th term is 54 and the 7th term is 1458.
Answer: We are given \( a_4 = 54 \) and \( a_7 = 1458 \).

Using the formula \( a_n = ar^{n-1} \):

\( \Rightarrow a_4 = a(r)^{4-1} \)

\( \Rightarrow 54 = ar^3 \quad \text{(Equation 1)} \)

\( \Rightarrow a_7 = a(r)^{7-1} \)

\( \Rightarrow 1458 = a(r)^6 \quad \text{(Equation 2)} \)

Dividing Equation 2 by Equation 1:

\( \Rightarrow \frac{ar^6}{ar^3} = \frac{1458}{54} \)

\( \Rightarrow r^3 = 27 \)

\( \Rightarrow r = \sqrt[3]{27} \)

\( \Rightarrow r = 3 \)

Substituting \( r = 3 \) into Equation 1:

\( \Rightarrow a(3)^3 = 54 \)

\( \Rightarrow 27a = 54 \)

\( \Rightarrow a = 2 \)

Now we can find all terms:

\( a_1 = a = 2 \)

\( a_2 = ar = 2 \times 3 = 6 \)

\( a_3 = ar^2 = 2(3)^2 = 2 \times 9 = 18 \)

\( a_4 = ar^3 = 2(3)^3 = 2 \times 27 = 54 \)

Therefore, the required G.P. is 2, 6, 18, 54, ...
In simple words: Divide the two given terms to find the common ratio, then use one term to find the first term. Once you have these, you can list the entire sequence.

Exam Tip: Always start by finding r using the given terms, then find a. This approach is cleaner and avoids solving quadratic equations unnecessarily.

Question 11. The sum of first three terms of a G.P. is \( \frac{39}{10} \) and their product is 1. Find the common ratio and the terms.
Answer: We are given \( S_3 = \frac{39}{10} \) and \( a_1 \times a_2 \times a_3 = 1 \).

Let the three numbers in G.P. be \( \frac{a}{r}, a, ar \).

From the product condition:

\( \frac{a}{r} \times a \times ar = 1 \)

\( \Rightarrow a^3 = 1 \)

\( \Rightarrow a = 1 \)

From the sum condition:

\( S_3 = \frac{39}{10} \)

\( \Rightarrow \frac{a}{r} + a + ar = \frac{39}{10} \)

\( \Rightarrow a \left( \frac{1}{r} + 1 + r \right) = \frac{39}{10} \)

\( \Rightarrow 1 \left( \frac{1 + r + r^2}{r} \right) = \frac{39}{10} \)

\( \Rightarrow 10(1 + r + r^2) = 39r \)

\( \Rightarrow 10 + 10r + 10r^2 = 39r \)

\( \Rightarrow 10r^2 + 10r - 39r + 10 = 0 \)

\( \Rightarrow 10r^2 - 29r + 10 = 0 \)

\( \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \)

\( \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \)

\( \Rightarrow (5r - 2)(2r - 5) = 0 \)

\( \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \)

\( \Rightarrow r = \frac{2}{5} \text{ or } r = \frac{5}{2} \)

For \( r = \frac{5}{2} \):

Terms are: \( \frac{a}{r} = \frac{1}{\frac{5}{2}} = \frac{2}{5}, \quad a = 1, \quad ar = 1 \times \frac{5}{2} = \frac{5}{2} \)

For \( r = \frac{2}{5} \):

Terms are: \( \frac{a}{r} = \frac{1}{\frac{2}{5}} = \frac{5}{2}, \quad a = 1, \quad ar = 1 \times \frac{2}{5} = \frac{2}{5} \)

Therefore, the common ratio is \( \frac{2}{5} \) or \( \frac{5}{2} \), and the terms are \( \frac{5}{2}, 1, \frac{2}{5} \) or \( \frac{2}{5}, 1, \frac{5}{2} \).
In simple words: From the product, we get the middle term is 1. Then use the sum to find the common ratio, which gives two answers. The terms are just these three numbers in different orders depending on which ratio you choose.

Exam Tip: Always use the product condition first to simplify - it often gives you the middle term directly. Then the sum condition becomes much easier to solve.

Question 12. Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.
Answer: Let the three numbers in A.P. be \( a - d, a, a + d \).

From the sum condition:

\( a - d + a + a + d = 15 \)

\( \Rightarrow 3a = 15 \)

\( \Rightarrow a = 5 \)

After adding 1, 4 and 19 respectively, the terms become:

\( a - d + 1, a + 4, a + d + 19 \)

\( \Rightarrow 5 - d + 1, 5 + 4, 5 + d + 19 \)

\( \Rightarrow 6 - d, 9, d + 24 \)

These form a G.P., so the middle term squared equals the product of the first and third terms:

\( \frac{9}{6 - d} = \frac{d + 24}{9} \)

\( \Rightarrow 9^2 = (6 - d)(d + 24) \)

\( \Rightarrow 81 = 6d - d^2 + 144 - 24d \)

\( \Rightarrow 81 = -d^2 - 18d + 144 \)

\( \Rightarrow d^2 + 18d - 63 = 0 \)

\( \Rightarrow d^2 + 21d - 3d - 63 = 0 \)

\( \Rightarrow d(d + 21) - 3(d + 21) = 0 \)

\( \Rightarrow (d - 3)(d + 21) = 0 \)

\( \Rightarrow d = 3 \text{ or } d = -21 \)

For \( d = 3 \):

\( a - d = 5 - 3 = 2, \quad a = 5, \quad a + d = 5 + 3 = 8 \)

For \( d = -21 \):

\( a - d = 5 - (-21) = 26, \quad a = 5, \quad a + d = 5 + (-21) = -16 \)

Therefore, the required numbers are 2, 5 and 8 or 26, 5 and -16.
In simple words: Find the middle term from the A.P. sum. Then after adding the given numbers, use the G.P. property to find the common difference. You get two possible sets of numbers.

Exam Tip: Check both solutions by verifying: (i) they are in A.P. with the given middle term, and (ii) after adding the respective constants, they form a G.P. This guards against arithmetic errors.

Exercise 9.5

Question 1. Find the sum of:
(i) 20 terms of the series 2 + 6 + 18 + ...
(ii) 10 terms of the series \( 1 + \sqrt{3} + 3 + \) ...
(iii) 6 terms of the G.P. \( 1, -\frac{2}{3}, \frac{4}{9}, \) ...
(iv) 5 terms and n terms of the series \( 1 + \frac{2}{3} + \frac{4}{9} + \) ...
Answer:
(i) The series 2 + 6 + 18 + ... is a G.P. with first term \( a = 2 \) and common ratio \( r = \frac{6}{2} = 3 \).

Using the formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( S_{20} = \frac{2(3^{20} - 1)}{3 - 1} \)

\( = \frac{2(3^{20} - 1)}{2} \)

\( = 3^{20} - 1 \)

Therefore, the sum of the series is \( 3^{20} - 1 \).

(ii) The series \( 1 + \sqrt{3} + 3 + \) ... is a G.P. with first term \( a = 1 \) and common ratio \( r = \sqrt{3} \).

Using the formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( S_{10} = \frac{1((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} \)

\( = \frac{(3)^{10/2} - 1}{\sqrt{3} - 1} \)

\( = \frac{3^5 - 1}{\sqrt{3} - 1} \)

\( = \frac{243 - 1}{\sqrt{3} - 1} \)

\( = \frac{242}{\sqrt{3} - 1} \)

Multiply numerator and denominator by \( \sqrt{3} + 1 \):

\( = \frac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)

\( = \frac{242(\sqrt{3} + 1)}{3 - 1} \)

\( = \frac{242(\sqrt{3} + 1)}{2} \)

\( = 121(\sqrt{3} + 1) \)

Therefore, the sum of the series is \( 121(\sqrt{3} + 1) \).

(iii) The G.P. \( 1, -\frac{2}{3}, \frac{4}{9}, \) ... has first term \( a = 1 \) and common ratio \( r = -\frac{2}{3} \).

Using the formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( S_6 = \frac{1 \left[ \left( -\frac{2}{3} \right)^6 - 1 \right]}{-\frac{2}{3} - 1} \)

\( = \frac{\frac{2^6}{3^6} - 1}{-\frac{2}{3} - 1} \)

\( = \frac{\frac{64}{729} - 1}{-\frac{2 + 3}{3}} \)

\( = \frac{\frac{64 - 729}{729}}{-\frac{5}{3}} \)

\( = \frac{\frac{-665}{729}}{-\frac{5}{3}} \)

\( = \frac{-665}{729} \times \frac{-3}{5} \)

\( = \frac{665 \times 3}{729 \times 5} \)

\( = \frac{1995}{3645} \)

\( = \frac{133}{243} \)

Therefore, the sum of the series is \( \frac{133}{243} \).

(iv) For the series \( 1 + \frac{2}{3} + \frac{4}{9} + \) ..., the first term is \( a = 1 \) and the common ratio is \( r = \frac{2}{3} \).

Sum of 5 terms:

\( S_5 = \frac{1 \left[ \left( \frac{2}{3} \right)^5 - 1 \right]}{\frac{2}{3} - 1} \)

\( = \frac{\frac{2^5}{3^5} - 1}{-\frac{1}{3}} \)

\( = \frac{\frac{32}{243} - 1}{-\frac{1}{3}} \)

\( = \frac{\frac{32 - 243}{243}}{-\frac{1}{3}} \)

\( = \frac{\frac{-211}{243}}{-\frac{1}{3}} \)

\( = \frac{-211}{243} \times \frac{-3}{1} \)

\( = \frac{211 \times 3}{243} \)

\( = \frac{633}{243} \)

\( = \frac{211}{81} \)

Therefore, the sum of 5 terms is \( \frac{211}{81} \).

Sum of n terms:

\( S_n = \frac{1 \left[ \left( \frac{2}{3} \right)^n - 1 \right]}{\frac{2}{3} - 1} \)

\( = \frac{\left( \frac{2}{3} \right)^n - 1}{-\frac{1}{3}} \)

\( = -3 \left[ \left( \frac{2}{3} \right)^n - 1 \right] \)

\( = 3 - 3 \left( \frac{2}{3} \right)^n \)

Therefore, the sum of n terms is \( 3 - 3 \left( \frac{2}{3} \right)^n \).
In simple words: For each G.P., identify the first term and common ratio. Apply the sum formula carefully. When the denominator has a square root, rationalize it by multiplying by the conjugate.

Exam Tip: Always simplify your final answer to the lowest terms. For series with irrational common ratios, rationalizing denominators is expected and scores marks for proper presentation.

Question 2. Find the sum of the series 81 - 27 + 9 - ..... - 1/27.
Answer: This series forms a G.P. where the first term is 81, and the common ratio equals -1/3. The last term is -1/27. To find the sum, we apply the formula \( S_n = \frac{a - lr}{1 - r} \), where \( a \) is the first term, \( l \) is the last term, and \( r \) is the common ratio.

\( S_n = \frac{81 - (-\frac{1}{27})(-\frac{1}{3})}{1 - (-\frac{1}{3})} \)

\( = \frac{81 - \frac{1}{81}}{1 + \frac{1}{3}} \)

\( = \frac{\frac{6561 - 1}{81}}{\frac{3 + 1}{3}} \)

\( = \frac{\frac{6560}{81}}{\frac{4}{3}} \)

\( = \frac{6560 \times 3}{81 \times 4} \)

\( = \frac{19680}{324} = \frac{1640}{27} \)

Therefore, the sum of the series is \( \frac{1640}{27} \).
In simple words: When you add all the terms in this series together, you get a fraction equal to 1640 divided by 27.

Exam Tip: Remember the sum formula for G.P. when the last term is given - always use \( S_n = \frac{a - lr}{1 - r} \) to avoid calculation errors with the common ratio.

Question 3. The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.
Answer: You are given that \( a_n = 128 \), \( r = 2 \), and \( S_n = 255 \). Using the formula \( a_n = ar^{n-1} \):

\( 128 = a(2)^{n-1} \)

\( a \cdot \frac{2^n}{2} = 128 \)

\( 2^n = \frac{256}{a} \) ... (Eq 1)

Now apply the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( 255 = \frac{a(2^n - 1)}{2 - 1} \)

\( 255 = a(2^n - 1) \)

Substitute the value of \( 2^n \) from Eq 1:

\( 255 = a\left(\frac{256}{a} - 1\right) \)

\( 255 = 256 - a \)

\( a = 256 - 255 = 1 \)

Therefore, the first term of the G.P. is 1.
In simple words: By using the given information about the last term and the sum, you can work backwards to find that the first term must equal 1.

Exam Tip: When you have both the nth term and the sum given, create two equations and substitute one into the other to eliminate the variable n.

Question 4(i). How many terms of the G.P. 3, 3^2, 3^3, .... are needed to give the sum 120?
Answer: The given series forms a G.P. with first term \( a = 3 \) and common ratio \( r = 3 \). Let \( n \) represent the number of terms required to achieve a sum of 120.

Using the sum formula:

\( 120 = \frac{3(3^n - 1)}{3 - 1} \)

\( 120 = \frac{3(3^n - 1)}{2} \)

\( 240 = 3(3^n - 1) \)

\( 80 = 3^n - 1 \)

\( 81 = 3^n \)

\( 3^4 = 3^n \)

\( n = 4 \)

Thus, 4 terms of this G.P. are required to produce a sum of 120.
In simple words: You need to add up the first 4 terms of this series to get a total of 120.

Exam Tip: Always simplify the sum formula step by step, and look for powers that match - here, recognizing that 81 = 3^4 makes finding n straightforward.

Question 4(ii). How many terms of the G.P. 1, 4, 16, ... must be taken to have their sum equal to 341?
Answer: The above series forms a G.P. where the first term is \( a = 1 \) and the common ratio is \( r = 4 \). Let \( n \) be the number of terms needed to achieve a sum of 341.

Using the sum formula:

\( 341 = \frac{1(4^n - 1)}{4 - 1} \)

\( 341 = \frac{4^n - 1}{3} \)

\( 1023 = 4^n - 1 \)

\( 4^n = 1024 \)

\( 4^n = 4^5 \)

\( n = 5 \)

Therefore, 5 terms of this G.P. must be added to obtain a sum of 341.
In simple words: When you add the first 5 numbers from this series - 1, then 4, then 16, then 64, then 256 - they total exactly 341.

Exam Tip: Pay attention to powers of the common ratio - converting 1024 to 4^5 is the key step that reveals your answer for n.

Question 5. How many terms of the series, \( \frac{2}{9} - \frac{1}{3} + \frac{1}{2} + \).... will make the sum \( \frac{55}{72} \)?
Answer: The series shown is a G.P. where the first term equals \( a = \frac{2}{9} \) and the common ratio is \( r = \frac{\frac{-1}{3}}{\frac{2}{9}} = -\frac{1}{3} \times \frac{9}{2} = -\frac{3}{2} \). Let \( n \) be the number of terms needed to obtain a sum of \( \frac{55}{72} \).

Using the sum formula:

\( \frac{55}{72} = \frac{\frac{2}{9}[(-\frac{3}{2})^n - 1]}{-\frac{3}{2} - 1} \)

\( \frac{55}{72} = \frac{\frac{2}{9}[(-\frac{3}{2})^n - 1]}{-\frac{5}{2}} \)

\( \frac{55}{72} = \frac{4[(-\frac{3}{2})^n - 1]}{9 \times (-5)} \)

\( \frac{55}{72} = \frac{4[(-\frac{3}{2})^n - 1]}{-45} \)

\( \frac{55 \times (-45)}{72 \times 4} = (-\frac{3}{2})^n - 1 \)

\( -\frac{2475}{288} = (-\frac{3}{2})^n - 1 \)

\( -\frac{2475}{288} + 1 = (-\frac{3}{2})^n \)

\( \frac{-2475 + 288}{288} = (-\frac{3}{2})^n \)

\( (-\frac{3}{2})^n = -\frac{2187}{288} \)

Dividing numerator and denominator by 9:

\( (-\frac{3}{2})^n = -\frac{243}{32} \)

\( (-\frac{3}{2})^n = (-\frac{3}{2})^5 \)

\( n = 5 \)

Therefore, 5 terms of this series are required to produce the sum \( \frac{55}{72} \).
In simple words: When you add the first 5 terms together in this alternating series, the total comes to exactly \( \frac{55}{72} \).

Exam Tip: With negative common ratios, take care during simplification - verifying that both sides match the same power base (here, both sides as powers of -3/2) confirms your answer.

Question 6. The 2nd and 5th terms of a geometric series are - 1/2 and 1/16 respectively. Find the sum of the series upto 8 terms.
Answer: You are given \( a_2 = -\frac{1}{2} \) and \( a_5 = \frac{1}{16} \). Using the formula \( a_n = ar^{n-1} \), you have \( a_2 = ar \) and \( a_5 = ar^4 \). Dividing these expressions:

\( \frac{ar^4}{ar} = \frac{\frac{1}{16}}{-\frac{1}{2}} \)

\( r^3 = -\frac{1}{8} \)

\( r^3 = (-\frac{1}{2})^3 \)

\( r = -\frac{1}{2} \)

Since \( a_2 = ar \):

\( -\frac{1}{2} = a(-\frac{1}{2}) \)

\( a = 1 \)

Now use the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( S_8 = \frac{1[(-\frac{1}{2})^8 - 1]}{-\frac{1}{2} - 1} \)

\( = \frac{\frac{1}{256} - 1}{-\frac{3}{2}} \)

\( = \frac{\frac{1 - 256}{256}}{-\frac{3}{2}} \)

\( = \frac{-\frac{255}{256}}{-\frac{3}{2}} \)

\( = \frac{-255 \times 2}{256 \times (-3)} \)

\( = \frac{-510}{-768} = \frac{510}{768} \)

Dividing numerator and denominator by 6:

\( S_8 = \frac{85}{128} \)

The sum of the series up to 8 terms is \( \frac{85}{128} \).
In simple words: First find the common ratio by dividing two known terms, then locate the first term, and finally apply the sum formula with all eight terms.

Exam Tip: When given two terms of a G.P. that are not consecutive, dividing them eliminates the first term and reveals a power of the common ratio directly.

Question 7. The first term of a G.P. is 27 and 8th term is 1/81. Find the sum of its first 10 terms.
Answer: You have \( a = 27 \) and \( a_8 = \frac{1}{81} \). Using the formula \( a_n = ar^{n-1} \):

\( a_8 = ar^7 \)

\( \frac{1}{81} = 27r^7 \)

\( r^7 = \frac{1}{81 \times 27} = \frac{1}{3^4 \times 3^3} = \frac{1}{3^7} \)

\( r = \frac{1}{3} \)

Now apply the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( S_{10} = \frac{27[(\frac{1}{3})^{10} - 1]}{\frac{1}{3} - 1} \)

\( = \frac{27(\frac{1}{3^{10}} - 1)}{-\frac{2}{3}} \)

\( = \frac{27 \times \frac{1 - 3^{10}}{3^{10}}}{-\frac{2}{3}} \)

\( = \frac{3^3(1 - 3^{10})}{3^{10}} \times \frac{-3}{2} \)

\( = \frac{3^4(1 - 3^{10})}{3^{10} \times (-2)} \)

\( = \frac{81}{2} \times \frac{-(1 - 3^{10})}{3^{10}} \)

\( = \frac{81}{2}(1 - \frac{1}{3^{10}}) \)

Therefore, the sum of the first 10 terms is \( \frac{81}{2}(1 - \frac{1}{3^{10}}) \).
In simple words: Once you identify the common ratio as 1/3, plug everything into the sum formula to get the total of the first 10 terms.

Exam Tip: When working with fractional common ratios, be careful with negative signs in the denominator - keep track of how they affect your final answer.

Question 8. Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.
Answer: You are given \( r = 3 \), \( l = a_n = 486 \), and \( S_n = 728 \). Using the formula \( a_n = ar^{n-1} \):

\( 486 = a(3)^{n-1} \)

\( a \cdot \frac{3^n}{3} = 486 \)

\( 3^n \cdot a = 1458 \)

\( a = \frac{1458}{3^n} \) ... (Eq 1)

Now apply the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( 728 = \frac{\frac{1458}{3^n}(3^n - 1)}{3 - 1} \)

\( 728 = \frac{1458 - \frac{1458}{3^n}}{2} \)

\( 1456 = 1458 - \frac{1458}{3^n} \)

\( \frac{1458}{3^n} = 1458 - 1456 = 2 \)

\( 3^n = \frac{1458}{2} = 729 \)

Substituting back into Eq 1:

\( a = \frac{1458}{729} = 2 \)

Therefore, the first term of the G.P. is 2.
In simple words: Using both the last term formula and the sum formula together, you can solve for the number of terms and then find that the first term equals 2.

Exam Tip: When you have expressions for a in terms of n from different formulas, set them equal and solve for n first, then substitute back to find a.

Question 9. In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.
Answer: Let the last term be the nth term of the G.P. You have \( a = 7 \), \( l = a_n = 448 \), and \( S_n = 889 \). Using the formula \( a_n = ar^{n-1} \):

\( 448 = 7(r)^{n-1} \)

\( r^{n-1} = 64 \)

\( \frac{r^n}{r} = 64 \)

\( r^n = 64r \) ... (Eq 1)

Now apply the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( 889 = \frac{7(64r - 1)}{r - 1} \)

\( 889(r - 1) = 448r - 7 \)

\( 889r - 889 = 448r - 7 \)

\( 889r - 448r = 889 - 7 \)

\( 441r = 882 \)

\( r = 2 \)

Therefore, the common ratio of the G.P. is 2.
In simple words: By setting up two equations from the given information and eliminating terms, you find that the common ratio must be 2.

Exam Tip: Express r^n in one form from the nth-term equation, then substitute it into the sum formula to create one equation with r as the only unknown.

Question 10. Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.
Answer: You are given \( r = 3 \) and \( S_7 = 2186 \). Using the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( 2186 = \frac{a(3^7 - 1)}{3 - 1} \)

\( 2186 = \frac{a(2187 - 1)}{2} \)

\( 2186 = \frac{2186a}{2} \)

\( a = 2 \)

Now find the third term using \( a_3 = ar^2 \):

\( a_3 = 2(3)^2 = 2(9) = 18 \)

Therefore, the third term of the G.P. is 18.
In simple words: First determine the first term from the sum formula, then multiply by the common ratio twice to get the third term.

Exam Tip: Notice that the coefficient in the numerator equals the given sum - this means the calculation simplifies directly to find a = 2.

Question 11. If the first term of a G.P. is 5 and the sum of first three terms is 31/5, find the common ratio.
Answer: You have \( a = 5 \) and \( S_3 = \frac{31}{5} \). Using the sum formula \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\( \frac{31}{5} = \frac{5(r^3 - 1)}{r - 1} \)

\( 31(r - 1) = 25(r^3 - 1) \)

Since \( r^3 - 1 = (r - 1)(r^2 + r + 1) \):

\( 31(r - 1) = 25(r - 1)(r^2 + r + 1) \)

Dividing both sides by \( (r - 1) \):

\( 31 = 25(r^2 + r + 1) \)

\( 31 = 25r^2 + 25r + 25 \)

\( 6 = 25r^2 + 25r \)

\( 25r^2 + 25r - 6 = 0 \)

Using the quadratic formula or factoring, this gives \( r = \frac{2}{5} \) or \( r = -\frac{3}{5} \).

Therefore, the common ratio is either \( \frac{2}{5} \) or \( -\frac{3}{5} \).
In simple words: Setting up the sum formula with three terms and the first term value, then factoring the cubic expression, reveals that two different common ratios can work.

Exam Tip: When you have a cubic equation, look for the factorization \( r^3 - 1 = (r - 1)(r^2 + r + 1) \) to simplify and reduce to a quadratic, which may have two valid solutions.

Question 12. In a Geometric Progression (G.P.) the first term is 24 and the fifth term is 8. Find the ninth term of the G.P.
Answer: Let the first term be a and the common ratio be r. We are given that a = 24 and ar⁴ = 8. Dividing the fifth term by the first term gives us:
\( \frac{ar^4}{a} = \frac{8}{24} \)
\( r^4 = \frac{1}{3} \)
\( (r^4)^2 = \left(\frac{1}{3}\right)^2 \)
\( r^8 = \frac{1}{9} \)
The ninth term is calculated as ar⁸:
\( a_9 = ar^8 = 24 \times \frac{1}{9} = \frac{8}{3} \)
In simple words: To find the ninth term, we first find r⁸ by using the information about the fifth term. Then we multiply the first term by r⁸ to get our answer.

Exam Tip: Always use the given information to find the common ratio raised to the required power - this avoids having to find r itself, which can involve messy roots.

Question 1. The 10th term of the A.P. 5, 8, 11, 14, .... is
(a) 32
(b) 35
(c) 38
(d) 185
Answer: (a) 32
In simple words: The sequence goes up by 3 each time. Starting from 5, after 9 steps of adding 3, we reach 32.

Exam Tip: Remember that the 10th term means we add the common difference 9 times, not 10 times - the formula \( a_n = a + (n-1)d \) accounts for this.

Question 2. The 30th term of the A.P. 10, 7, 4, ... is
(a) 87
(b) 77
(c) -77
(d) -87
Answer: (c) -77
In simple words: The series decreases by 3 each time. After 29 steps of subtracting 3 from 10, the result becomes negative.

Exam Tip: Watch the sign of the common difference - a negative common difference means the sequence is decreasing, leading to negative terms as we move further along.

Question 3. The 11th term of the A.P. \( -3, -\frac{1}{2}, 2, ... \) is
(a) 28
(b) 22
(c) -38
(d) \( -48\frac{1}{2} \)
Answer: (b) 22
In simple words: The first term is -3. Each step adds \( \frac{5}{2} \). After 10 steps, we add \( 10 \times \frac{5}{2} = 25 \) to the starting value, giving us 22.

Exam Tip: When working with fractional common differences, convert everything to fractions or decimals to avoid arithmetic errors.

Question 4. The 15th term from the last of the A.P. 7, 10, 13, ... , 130 is
(a) 49
(b) 85
(c) 88
(d) 110
Answer: (c) 88
In simple words: When counting from the end, 130 is the last term. The 15th position back from the end is found by subtracting 14 times the common difference (3) from 130.

Exam Tip: For "nth term from the last," use the formula: nth term from end = l - (n-1)d, where l is the last term.

Question 5. If the common difference of the A.P. is 5, then a₁₈ - a₁₃ is
(a) 5
(b) 20
(c) 25
(d) 30
Answer: (c) 25
In simple words: The difference between two terms in an A.P. equals the common difference times the gap between the positions. Here, the gap is 18 - 13 = 5, so the answer is 5 × 5 = 25.

Exam Tip: Notice that a₁₈ - a₁₃ = (18 - 13)d = 5d - you don't need to find the individual terms.

Question 6. In an A.P., if a₁₈ - a₁₄ = 32 then the common difference is
(a) 8
(b) -8
(c) -4
(d) 4
Answer: (a) 8
In simple words: The difference between two terms equals the gap in positions times d. So 32 = (18 - 14)d = 4d, meaning d = 8.

Exam Tip: Work backwards from the difference between terms - divide by the position gap to find the common difference.

Question 7. In an A.P., if d = -4, n = 7, aₙ = 4, then a is
(a) 6
(b) 7
(c) 20
(d) 28
Answer: (d) 28
In simple words: We substitute into the formula: 4 = a + (7 - 1)(-4), so 4 = a - 24, giving a = 28.

Exam Tip: Rearrange the formula \( a_n = a + (n-1)d \) to solve for a when the other values are known.

Question 8. In an A.P., if a = 3.5, d = 0, n = 101, then aₙ will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Answer: (b) 3.5
In simple words: When the common difference is 0, every term in the sequence equals the first term. So a₁₀₁ = 3.5.

Exam Tip: A common difference of 0 creates a constant sequence - all terms are identical to the first term.

Question 9. Which term of the A.P. 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Answer: (b) 10th
In simple words: We set aₙ = 210 and solve: 210 = 21 + (n - 1)(21). This gives 21n = 210, so n = 10.

Exam Tip: To find which term equals a given value, set up the equation aₙ = value and solve for n.

Question 10. If the last term of A.P. 5, 3, 1, -1, .... is -41, then the A.P. consists of
(a) 46 terms
(b) 25 terms
(c) 24 terms
(d) 23 terms
Answer: (c) 24 terms
In simple words: The first term is 5 and the common difference is -2. Setting aₙ = -41 and solving gives n = 24.

Exam Tip: When finding the number of terms, solve for n in the formula \( a_n = a + (n-1)d \) using the last term value.

Question 11. If k - 1, k + 1 and 2k + 3 are in A.P. , then the value of k is
(a) -2
(b) 0
(c) 2
(d) 4
Answer: (b) 0
In simple words: In an A.P., the difference between consecutive terms is constant. So (k + 1) - (k - 1) must equal (2k + 3) - (k + 1). This gives 2 = k + 2, so k = 0.

Exam Tip: For three terms in A.P., use: second term - first term = third term - second term, or equivalently, the middle term is the average of the other two.

Question 12. The 21st term of an A.P. whose first two terms are -3 and 4 is
(a) 17
(b) 137
(c) 143
(d) -143
Answer: (b) 137
In simple words: From the first two terms, the common difference is 4 - (-3) = 7. Then a₂₁ = -3 + (21 - 1)(7) = -3 + 140 = 137.

Exam Tip: When given the first two terms, subtract the first from the second to find d, then apply the formula for the required term.

Question 13. If the first term of an A.P. is -5 and the common difference is 2, then the sum of its first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Answer: (a) 0
In simple words: Using the sum formula, \( S_6 = \frac{6}{2}[2(-5) + (6-1)(2)] = 3[-10 + 10] = 0 \).

Exam Tip: The sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \) is faster than adding all terms individually, especially for larger values of n.

Question 14. The sum of 25 terms of the A.P., \( -\frac{2}{3}, -\frac{2}{3}, -\frac{2}{3}, ... \) is
(a) 0
(b) \( -\frac{2}{3} \)
(c) \( -\frac{50}{3} \)
(d) -50
Answer: (c) \( -\frac{50}{3} \)
In simple words: This A.P. has first term \( -\frac{2}{3} \) and common difference 0. The sum is \( S_{25} = \frac{25}{2}[2 \times (-\frac{2}{3}) + 0] = \frac{25}{2} \times (-\frac{4}{3}) = -\frac{50}{3} \).

Exam Tip: When d = 0, every term equals the first term, so the sum is simply \( S_n = n \times a \).

Question 15. In an A.P. if a = 1, aₙ = 20 and Sₙ = 399, then n is
(a) 19
(b) 21
(c) 38
(d) 42
Answer: (c) 38
In simple words: From aₙ = 20, we get (n - 1)d = 19, so d = \( \frac{19}{n-1} \). Substituting into the sum formula: \( 399 = \frac{n}{2}[2 + 19] = \frac{n}{2} \times 21 \). Solving gives n = 38.

Exam Tip: When both the last term and sum are given, express d in terms of n using the last term formula, then substitute into the sum formula.

Question 16. In an A.P., if a = -5, l = 21 and S = 200, then n is equal to
(a) 50
(b) 40
(c) 32
(d) 25
Answer: (d) 25
In simple words: There is an alternate sum formula: \( S_n = \frac{n}{2}(a + l) \), where l is the last term. Substituting: \( 200 = \frac{n}{2}(-5 + 21) = \frac{n}{2} \times 16 = 8n \). So n = 25.

Exam Tip: When the first and last terms are known, use \( S_n = \frac{n}{2}(a + l) \) instead of the standard formula - it's much quicker.

Question 17. The sum of first five multiples of 3 is
(1) 45
(2) 55
(3) 65
(4) 75
Answer: (1) 45
In simple words: The first five multiples of 3 make an arithmetic sequence: 3, 6, 9, 12, 15. When you add them up, you get 45.

Exam Tip: Always check if the given numbers form an arithmetic progression before applying the sum formula - this saves time and reduces calculation errors.

Question 18. The number of two digit numbers which are divisible by 3 is
(1) 33
(2) 31
(3) 30
(4) 29
Answer: (3) 30
In simple words: Two-digit numbers divisible by 3 start at 12 and end at 99, forming the pattern 12, 15, 18,... 99. Using the formula for finding how many terms are in this sequence, we get 30 such numbers.

Exam Tip: When counting terms in an arithmetic sequence, always identify the first term, last term, and common difference correctly before using the formula \( a_n = a + (n-1)d \).

Question 19. The number of multiples of 4 that lie between 10 and 250 is
(1) 62
(2) 60
(3) 59
(4) 55
Answer: (2) 60
In simple words: The multiples of 4 between 10 and 250 are 12, 16, 20,... up to 248. They form an arithmetic sequence with first term 12, last term 248, and difference 4. This sequence has 60 terms.

Exam Tip: Always be careful with "between" - check whether the boundary numbers themselves are multiples, and use the correct last term that falls within the range.

Question 20. The sum of first 10 even whole numbers is
(1) 110
(2) 90
(3) 55
(4) 45
Answer: (2) 90
In simple words: The first 10 even whole numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18. They form an arithmetic sequence starting at 0 and ending at 18. Their total is 90.

Exam Tip: Remember that even whole numbers start from 0, not from 2. Use the sum formula \( S_n = \frac{n}{2}(a + l) \) where a is the first term and l is the last term.

Question 21. The 11th term of the G.P. \( \frac{1}{8}, - \frac{1}{4}, 2, -1, \ldots \) is
(1) 64
(2) - 64
(3) 128
(4) - 128
Answer: (3) 128
In simple words: This is a geometric sequence (G.P.) where each term is found by multiplying the previous term by - 2. Starting from \( \frac{1}{8} \), the 11th term works out to 128.

Exam Tip: In a G.P., always find the common ratio r by dividing any term by the previous one, then use \( a_n = ar^{n-1} \) carefully - watch the signs when r is negative.

Question 22. The 5th term from the end of the G.P. 2, 6, 18, \ldots, 13122 is
(1) 162
(2) 486
(3) 54
(4) 1458
Answer: (1) 162
In simple words: To find a term from the end, work backwards using the formula for the nth term from the end: \( l \left( \frac{1}{r} \right)^{n-1} \), where l is the last term. The 5th term from the end equals 162.

Exam Tip: When finding terms from the end of a G.P., treat the last term as your starting point and use the reciprocal of the common ratio to count backwards.

Question 23. If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is
(1) - 1
(2) - 4
(3) 1
(4) 4
Answer: (2) - 4
In simple words: In a geometric sequence, the ratio between any two consecutive terms must be the same. Setting up this condition and solving gives k = - 4. We must reject k = - 1 because it makes the first term zero, which breaks the G.P. property.

Exam Tip: For three consecutive G.P. terms, use the property: (middle term)² = (first term) × (third term). Always verify your solutions don't make any term zero or undefined.

Question. Assertion (A): They are consecutive term of an arithmetic progression.
Reason (R): Difference between two successive term is - 3.
Given is a sequence of three terms: - 1, - 5, - 9,......
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (1) Assertion (A) is true, but Reason (R) is false.
In simple words: The sequence - 1, - 5, - 9 is indeed an arithmetic progression. However, the difference between consecutive terms is - 4, not - 3, so Reason (R) states the wrong difference value.

Exam Tip: In Assertion-Reason questions, both parts must be true AND the reason must correctly explain the assertion for the answer to be option (3). If the assertion is correct but the reason's explanation is wrong, choose option (1).

Question 2. Assertion (A): This A.P. will have all terms negative after 2nd term. Reason (R): The common difference is negative. Given, Tn = 4 - 2n is the nth term of an A.P.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The sequence is 2, 0, - 2, - 4,... The assertion is true because after the 2nd term (which is 0), all remaining terms (- 2, - 4,...) are negative. The reason is also true: the common difference is - 2, which is negative. A negative common difference causes the terms to decrease and eventually become negative.

Exam Tip: Calculate the first few terms using the formula before deciding whether the assertion is correct. Check the actual common difference value rather than assuming - it might not match what the reason states.

Question 3. Assertion (A): 53 is a term of this A.P. Reason (R): Its first term is 3 and common difference is 5. Observe the sequence of the terms: 3, 8, 13,................
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The sequence has first term 3 and common difference 5. Using the formula, when we set the nth term equal to 53, we get n = 11, so 53 is indeed the 11th term. Both the assertion and reason are correct, and the reason directly supports the assertion.

Exam Tip: Always verify that a given number is actually a term by checking whether n is a positive integer after solving \( a_n = a + (n-1)d \). If n comes out to be negative or a fraction, the number is not a term of the sequence.

Question 4. Assertion (A): z, y, x are in A.P. Reason (R): Terms of an A.P. taken in reverse order also forms an A.P. Given x, y, z are in A.P.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: If x, y, z are in arithmetic progression with common difference d, then when reversed to z, y, x, they still form an A.P. - but now with common difference - d (negative of the original). Both statements are true, and the reason fully explains why the assertion holds.

Exam Tip: Remember that reversing an A.P. gives another A.P. with the opposite common difference. Both the original and reversed sequences satisfy the A.P. property.

Question 5. Assertion (A): The sum of first 99 natural numbers is 4950. Reason (R): The sum of first n natural numbers is \( \frac{n(n + 1)}{2} \).
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The sum of the first 99 natural numbers (1 + 2 + 3 +... + 99) can be found using the formula \( \frac{99 \times 100}{2} = 4950 \). The formula stated in the reason correctly gives this result, and it is the proper way to justify the assertion.

Exam Tip: When verifying the sum formula, always substitute the value of n and compute carefully. For n = 99, multiply 99 × 100 = 9900, then divide by 2 to get 4950.

Chapter Test

Question 1. Write the first four terms of the A.P. when its first term is -5 and common difference is -3.
Answer: Start with a = -5 and d = -3. Using the formula for the nth term, you can find each successive term by adding the common difference to the previous term.
\( a_2 = -5 + (2-1) \times (-3) = -5 - 3 = -8 \)
\( a_3 = -5 + (3-1) \times (-3) = -5 - 6 = -11 \)
\( a_4 = -5 + (4-1) \times (-3) = -5 - 9 = -14 \)
The sequence runs: -5, -8, -11, -14.
In simple words: Begin with -5. Each time, subtract 3 to get the next number. So you get -5, then -8, then -11, then -14.

Exam Tip: Always double-check your first and second terms to ensure the common difference is calculated correctly before computing the rest.

Question 2. Verify that each of the following lists of numbers is an A.P., and then write its next three terms:
(i) \( 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots \)
(ii) \( 5, \frac{14}{3}, \frac{13}{3}, 4, \ldots \)
Answer:
(i) Check the differences between consecutive terms:
\( a_2 - a_1 = \frac{1}{4} - 0 = \frac{1}{4} \)
\( a_3 - a_2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \)
Since each term minus its predecessor equals \( \frac{1}{4} \), this forms an A.P. with common difference \( d = \frac{1}{4} \).
The next three terms:
\( a_5 = \frac{3}{4} + \frac{1}{4} = 1 \)
\( a_6 = 1 + \frac{1}{4} = \frac{5}{4} \)
\( a_7 = \frac{5}{4} + \frac{1}{4} = \frac{3}{2} \)
Next three terms: \( 1, \frac{5}{4}, \frac{3}{2} \)

(ii) Check the differences:
\( a_2 - a_1 = \frac{14}{3} - 5 = \frac{14-15}{3} = -\frac{1}{3} \)
\( a_3 - a_2 = \frac{13}{3} - \frac{14}{3} = -\frac{1}{3} \)
This is an A.P. with common difference \( d = -\frac{1}{3} \).
The next three terms:
\( a_5 = 4 + \left(-\frac{1}{3}\right) = \frac{11}{3} \)
\( a_6 = \frac{11}{3} - \frac{1}{3} = \frac{10}{3} \)
\( a_7 = \frac{10}{3} - \frac{1}{3} = 3 \)
Next three terms: \( \frac{11}{3}, \frac{10}{3}, 3 \)
In simple words: To verify an A.P., subtract each term from the next one. If you always get the same answer, it is an A.P. Then keep adding the common difference to get the new terms.

Exam Tip: Show all fraction calculations explicitly - examiners check that you can simplify correctly and work with common denominators.

Question 3. The nth term of an A.P. is 6n + 2. Find the common difference.
Answer: Given \( a_n = 6n + 2 \). To find the common difference, calculate the first two terms.
\( a_1 = 6(1) + 2 = 8 \)
\( a_2 = 6(2) + 2 = 14 \)
The common difference is \( d = a_2 - a_1 = 14 - 8 = 6 \).
In simple words: Plug in n = 1 and n = 2 into the formula. Subtract the first result from the second - that difference is d.

Exam Tip: When given a formula for the nth term, always find at least the first two terms to extract the common difference - never try to guess it from the formula itself.

Question 4. Show that the list of numbers 9, 12, 15, 18, ... form an A.P. Find its 16th term and the nth term.
Answer: Check that consecutive differences are constant:
\( a_2 - a_1 = 12 - 9 = 3 \)
\( a_3 - a_2 = 15 - 12 = 3 \)
Since the difference between any two adjacent terms is always 3, the list forms an A.P. with first term a = 9 and common difference d = 3.
Using the nth term formula \( a_n = a + (n-1)d \):
\( a_n = 9 + (n-1) \times 3 = 9 + 3n - 3 = 3n + 6 \)
For the 16th term:
\( a_{16} = 3(16) + 6 = 48 + 6 = 54 \)
Therefore, \( a_n = 3n + 6 \) and \( a_{16} = 54 \).
In simple words: When each number is 3 more than the one before, you have an A.P. The 16th term is 54, and any term can be found using 3n + 6.

Exam Tip: Always verify the A.P. property before finding terms - showing the constant difference proves it is an A.P., which is required for full marks.

Question 5. Find the 6th term from the end of the A.P. 17, 14, 11, ..., -40.
Answer: Given: first term a = 17, last term l = -40, and common difference d = 14 - 17 = -3.
To find the nth term from the end, use the formula: \( \text{nth term from end} = l - (n-1)d \)
\( 6\text{th term from end} = -40 - (6-1) \times (-3) = -40 - 5 \times (-3) = -40 + 15 = -25 \)
The 6th term from the end is -25.
In simple words: Start at the last number, -40. Go backwards 5 steps, where each backwards step means adding 3 (since d is negative). This gives you -40 + 15 = -25.

Exam Tip: Be careful with negative common differences when working backwards - the signs will flip, so subtracting a negative becomes adding.

Question 6. If the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P.
Answer: Given: \( a_8 = 31 \) and \( a_{15} - a_{11} = 16 \).
Using \( a_n = a + (n-1)d \):
\( a_8 = a + 7d \) ... (Eq 1)
\( a_{15} = a + 14d \) and \( a_{11} = a + 10d \)
From the condition \( a_{15} - a_{11} = 16 \):
\( (a + 14d) - (a + 10d) = 16 \)
\( 4d = 16 \)
\( d = 4 \)
Substitute into Eq 1:
\( a + 7(4) = 31 \)
\( a + 28 = 31 \)
\( a = 3 \)
Now find the first few terms:
\( a_1 = 3 \)
\( a_2 = 3 + 4 = 7 \)
\( a_3 = 7 + 4 = 11 \)
\( a_4 = 11 + 4 = 15 \)
The A.P. is 3, 7, 11, 15, ....
In simple words: Use the two given facts to make two equations. Solve them to find a and d, then list out the first few terms of the sequence.

Exam Tip: When a condition compares two terms (like "the 15th is 16 more than the 11th"), subtract the two term formulas to eliminate a - this leaves an equation in d only.

Question 7. The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the nth term.
Answer: Use \( a_n = a + (n-1)d \). From the condition \( a_{17} = 2(a_8) + 5 \):
\( a + 16d = 2(a + 7d) + 5 \)
\( a + 16d = 2a + 14d + 5 \)
\( 16d - 14d = 2a - a + 5 \)
\( 2d = a + 5 \)
\( a = 2d - 5 \) ... (Eq 1)
Given \( a_{11} = 43 \):
\( a + 10d = 43 \)
Substitute Eq 1:
\( (2d - 5) + 10d = 43 \)
\( 12d = 48 \)
\( d = 4 \)
From Eq 1: \( a = 2(4) - 5 = 3 \)
Therefore: \( a_n = 3 + (n-1) \times 4 = 3 + 4n - 4 = 4n - 1 \)
In simple words: Turn the relationship between the 17th and 8th terms into an equation. Combine it with the fact about the 11th term to find a and d. Then write the formula for the nth term.

Exam Tip: Always isolate one variable in terms of the other from the first condition, then substitute into the second condition - this reduces the system to a single equation.

Question 8. The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.
Answer: Use \( a_n = a + (n-1)d \). From \( a_{19} = 3(a_6) \):
\( a + 18d = 3(a + 5d) \)
\( a + 18d = 3a + 15d \)
\( 18d - 15d = 3a - a \)
\( 3d = 2a \)
\( a = \frac{3d}{2} \) ... (Eq 1)
Given \( a_9 = 19 \):
\( a + 8d = 19 \)
Substitute Eq 1:
\( \frac{3d}{2} + 8d = 19 \)
\( \frac{3d + 16d}{2} = 19 \)
\( \frac{19d}{2} = 19 \)
\( d = 2 \)
From Eq 1: \( a = \frac{3(2)}{2} = 3 \)
The first four terms:
\( a_1 = 3 \)
\( a_2 = 3 + 2 = 5 \)
\( a_3 = 5 + 2 = 7 \)
\( a_4 = 7 + 2 = 9 \)
The A.P. is 3, 5, 7, 9, ....
In simple words: Express a in terms of d using the first condition. Plug this into the equation for the 9th term. Solve for d, then find a. List the first few numbers.

Exam Tip: When fractions appear, multiply through by the denominator early to avoid carrying fractions through multiple steps.

Question 9. If the 3rd and the 9th terms of an A.P. are 4 and -8 respectively, then which term of the A.P. is zero?
Answer: Use \( a_n = a + (n-1)d \). Given \( a_3 = 4 \) and \( a_9 = -8 \):
\( a + 2d = 4 \) ... (1)
\( a + 8d = -8 \) ... (2)
Subtract equation (1) from equation (2):
\( 6d = -12 \)
\( d = -2 \)
From equation (1):
\( a + 2(-2) = 4 \)
\( a - 4 = 4 \)
\( a = 8 \)
To find which term equals zero, set \( a_n = 0 \):
\( 8 + (n-1)(-2) = 0 \)
\( 8 - 2n + 2 = 0 \)
\( 10 - 2n = 0 \)
\( n = 5 \)
The 5th term is zero.
In simple words: Set up two equations using the given information. Subtract to find d. Use d to find a. Then set a term equal to zero and solve for which position n gives that zero value.

Exam Tip: Always subtract equations strategically to eliminate one variable - choose the subtraction direction that makes the arithmetic simplest.

Question 10. Which term of the list of the numbers 5, 2, -1, -4, ... is -55?
Answer: This sequence is an A.P. with first term a = 5 and common difference d = 2 - 5 = -3.
Let the nth term equal -55, so \( a_n = -55 \).
Using the nth term formula:
\( -55 = 5 + (n-1)(-3) \)
\( -55 = 5 - 3n + 3 \)
\( -55 = 8 - 3n \)
\( 3n = 8 + 55 \)
\( 3n = 63 \)
\( n = 21 \)
The 21st term is -55.
In simple words: Identify the starting value and the step size (how much to subtract each time). Set up an equation where a term equals -55. Solve for n.

Exam Tip: When rearranging to isolate n, move the constant term first, then divide by the coefficient of n to avoid sign errors.

Question 11. The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term.
Answer: Use \( a_n = a + (n-1)d \). Given \( a_{24} = 2(a_{10}) \):
\( a + 23d = 2(a + 9d) \)
\( a + 23d = 2a + 18d \)
\( 23d - 18d = 2a - a \)
\( 5d = a \)
Now find the 72nd and 15th terms using \( a = 5d \):
\( a_{72} = a + 71d = 5d + 71d = 76d \)
\( a_{15} = a + 14d = 5d + 14d = 19d \)
Calculate four times the 15th term:
\( 4a_{15} = 4 \times 19d = 76d = a_{72} \)
Therefore, \( a_{72} = 4a_{15} \), which means the 72nd term is four times the 15th term. Proved.
In simple words: Use the given fact to find a relationship between a and d. Then substitute this into the formulas for the 72nd and 15th terms to show they match the required ratio.

Exam Tip: In "show that" problems, derive the simplest relationship between a and d from the given condition, then use it to verify the required result - avoid unnecessary calculations.

Question 12. Which term of the list of the numbers \( 20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \ldots \) is the first negative term?
Answer: Convert to improper fractions: the sequence is \( 20, \frac{77}{4}, \frac{37}{2}, \frac{71}{4}, \ldots \)
This is an A.P. with \( a = 20 \) and \( d = 19\frac{1}{4} - 20 = \frac{77}{4} - 20 = \frac{77-80}{4} = -\frac{3}{4} \).
The general term is:
\( a_n = 20 + (n-1) \times \left(-\frac{3}{4}\right) = 20 - \frac{3n}{4} + \frac{3}{4} = \frac{80 + 3}{4} - \frac{3n}{4} = \frac{83 - 3n}{4} \)
For the first negative term, \( a_n < 0 \):
\( \frac{83 - 3n}{4} < 0 \)
\( 83 - 3n < 0 \)
\( 3n > 83 \)
\( n > \frac{83}{3} = 27\frac{2}{3} \)
Since n must be a positive integer, the smallest value is \( n = 28 \).
The 28th term is the first negative term.
In simple words: Write the nth term formula. Set it less than zero. Solve for n. The smallest whole number n that satisfies this gives you the position of the first negative term.

Exam Tip: When working with inequalities and fractional common differences, clear fractions early by multiplying through, then be careful with inequality direction when dividing by a negative.

Question 13. How many three digit numbers are divisible by 9?
Answer: The smallest three-digit number divisible by 9 is 108, and the largest is 999. These form an A.P.: 108, 117, 126, ..., 999.
Here, a = 108, d = 9, and \( a_n = 999 \) (the last term).
Using the nth term formula:
\( a_n = a + (n-1)d \)
\( 999 = 108 + (n-1) \times 9 \)
\( 999 = 108 + 9n - 9 \)
\( 999 = 99 + 9n \)
\( 9n = 900 \)
\( n = 100 \)
There are 100 three-digit numbers divisible by 9.
In simple words: Find the smallest and largest three-digit multiples of 9. They form an A.P. Use the formula to count how many terms are in this sequence.

Exam Tip: Always identify the first and last terms of the sequence clearly before applying the formula - this is the most common source of error in counting problems.

Question 14. The sum of three numbers in A.P. is -3 and the product is 8. Find the numbers.
Answer: Let the three numbers in A.P. be \( a - d, a, a + d \).
Given their sum is -3:
\( (a - d) + a + (a + d) = -3 \)
\( 3a = -3 \)
\( a = -1 \)
Given their product is 8:
\( (a-d) \times a \times (a+d) = 8 \)
Substitute \( a = -1 \):
\( (-1-d)(-1)(-1+d) = 8 \)
\( (-1-d)(d-1)(-1) = 8 \)
Simplify:
\( (1+d)(d-1) = 8 \)
\( d^2 - 1 = 8 \)
\( d^2 = 9 \)
\( d = \pm 3 \)
If \( d = 3 \): the three numbers are \( -1-3 = -4, -1, -1+3 = 2 \).
If \( d = -3 \): the three numbers are \( -1+3 = 2, -1, -1-3 = -4 \).
Either way, the three numbers are -4, -1, and 2.
In simple words: Assume the form a - d, a, a + d for the three terms. Use the sum to find a. Use the product to find d. This gives you the numbers.

Exam Tip: The symmetric form \( a - d, a, a + d \) is perfect for three-term A.P. problems because the sum automatically simplifies to \( 3a \), making one equation trivial to solve.

Question 15. The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.
Answer: Let the angles be a, (a + d), (a + 2d), (a + 3d). According to the given condition, the largest angle equals twice the smallest angle, so (a + 3d) = 2a. This gives a = 3d. Substituting this back, the angles become 3d, 4d, 5d, 6d. Since the sum of angles in any quadrilateral is 360°, we have 3d + 4d + 5d + 6d = 360°, which means 18d = 360°. Therefore, d = 20°. The four angles are 60°, 80°, 100°, 120°.
In simple words: The four angles are 60 degrees, 80 degrees, 100 degrees, and 120 degrees. Each angle is 20 degrees more than the one before it.

Exam Tip: Always use the property that angles in a quadrilateral sum to 360°. Verify your answer by checking that the largest angle is indeed double the smallest.

Question 16. Find the sum of first 20 terms of an A.P. whose nth term is 15 - 4n.
Answer: Using the formula for the nth term, we calculate a₁ = 15 - 4(1) = 11 and a₂₀ = 15 - 4(20) = -65. To find the sum, we use the formula S_n = (n/2)[a + l], where a is the first term, l is the last term, and n is the number of terms. Substituting, S₂₀ = (20/2)[11 + (-65)] = 10(-54) = -540. Therefore, the sum of the first 20 terms is -540.
In simple words: The first term is 11 and the last (20th) term is -65. Adding them and multiplying by half the number of terms gives -540.

Exam Tip: When the nth term formula is given, always compute the first and last terms directly before applying the sum formula. This is faster than finding the common difference.

Question 17. Ten students of a class are lined up according to their heights. Height of first student is 150 cm, and every next student is 2 cm more in height. What is the total sum of heights of these ten students?
Answer: We have 10 students, with the first student's height a = 150 cm and the common difference d = 2 cm. Since each student is 2 cm taller than the previous one, the heights form an A.P. Using the A.P. sum formula S_n = (n/2)[2a + (n-1)d], we get S₁₀ = (10/2)[2(150) + (10-1)(2)] = 5[300 + 18] = 5(318) = 1590 cm. The total sum of heights of all ten students is 1590 cm.
In simple words: The heights are 150, 152, 154... and go up by 2 cm each time. Adding all 10 heights gives 1590 cm total.

Exam Tip: Recognize the real-world context as an A.P. problem. The sum formula with 2a and (n-1)d is usually faster than finding the last term separately when both the first term and common difference are given.

Question 18. Find the geometric progression whose 4th term is 54 and 7th term is 1458.
Answer: Let the first term be a and common ratio be r. We know a₄ = ar³ = 54 and a₇ = ar⁶ = 1458. Dividing a₇ by a₄ gives (ar⁶)/(ar³) = 1458/54, so r³ = 27. Taking the cube root, r = 3. Substituting r = 3 into ar³ = 54, we get a(3)³ = 54, so 27a = 54, giving a = 2. The terms of the G.P. are a₂ = ar = 2(3) = 6, a₃ = ar² = 2(9) = 18, a₄ = ar³ = 2(27) = 54. Therefore, the G.P. is 2, 6, 18, 54, ....
In simple words: Each term is 3 times the previous term. The first term is 2, and the sequence is 2, 6, 18, 54, and so on.

Exam Tip: Dividing two terms of a G.P. cancels the first term and gives you the common ratio directly. Always verify by checking that both given terms match your found values.

Question 19. The fourth term of a G.P. is the square of its second term and the first term is -3. Find its 7th term.
Answer: Given a = -3 and a₄ = (a₂)². We know a₄ = ar³ and a₂ = ar. From the condition ar³ = (ar)², we get (-3)r³ = (-3r)². This simplifies to -3r³ = 9r². Rearranging, 9r² + 3r³ = 0, so r²(9 + 3r) = 0. This gives r = 0 or r = -3. Since the common ratio cannot be zero, r = -3. Now, a₇ = ar⁶ = (-3)(-3)⁶ = -3 × 729 = -2187. Therefore, the 7th term is -2187.
In simple words: The common ratio is -3. Starting from -3, you multiply by -3 each time. The 7th term works out to -2187.

Exam Tip: When a condition like "a₄ = (a₂)²" is given, expand it using the general term formula immediately. Always reject r = 0 as it makes the sequence trivial. Verify your answer by checking the original condition.

Question 20. If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y, z are in G.P.
Answer: Let a₄ = x, a₁₀ = y, and a₁₆ = z. Using the general term formula, a₄ = ar³ = x, a₁₀ = ar⁹ = y, and a₁₆ = ar¹⁵ = z. Now, divide y by x: (y/x) = (ar⁹)/(ar³) = r⁶. Next, divide z by y: (z/y) = (ar¹⁵)/(ar⁹) = r⁶. Since (y/x) = (z/y) = r⁶, we have (y/x) = (z/y), which means y² = xz. This is the defining property of three terms in G.P. Therefore, x, y, z are in G.P.
In simple words: When you divide y by x, you get the same number as when you divide z by y. This means x, y, z follow a geometric pattern with a constant ratio.

Exam Tip: The key to proving three terms are in G.P. is showing that the ratio between consecutive terms is constant. Always reduce ratios of general terms first before substituting the given values.

Question 21. How many terms of the G.P. 3, (3/2), (3/4), .... are needed to give the sum (3069/512)?
Answer: For this G.P., a = 3 and r = (3/2)/3 = 1/2. We need to find n such that S_n = 3069/512. Using the sum formula S_n = a(r^n - 1)/(r - 1), we get 3069/512 = 3[(1/2)^n - 1]/(1/2 - 1). Simplifying, 3069/512 = 3[(1/2)^n - 1]/(-1/2) = -6[(1/2)^n - 1]. This gives 3069/512 = 1 - (1/2)^n. Rearranging, (1/2)^n = 1 - 3069/3072 = 3/3072. Simplifying further, (1/2)^n = 1/1024 = (1/2)^10. Therefore, n = 10. Hence, 10 terms are required to give the specified sum.
In simple words: You need to add up 10 terms of this series to reach the sum 3069/512.

Exam Tip: With fractional common ratios (r < 1), the sum formula converges quickly. Always express fractional powers as powers of 2 to solve for n easily. Double-check by verifying the arithmetic in simplifying the fractions.

Question 22. 15, 30, 60, 120 ...... are in G.P. (Geometric Progression).
(a) Find the nth term of this G.P. in terms of n.
Answer: For the G.P. 15, 30, 60, 120, the first term a = 15 and the common ratio r = 30/15 = 2. Using the general term formula a_n = ar^(n-1), we get a_n = 15 × 2^(n-1) = (15/2) × 2^n = 7.5 × 2^n. Therefore, the nth term is 7.5 × 2^n.
In simple words: Each term is 2 times the previous one, starting from 15. The nth term formula is 7.5 times 2 to the power n.

Exam Tip: Simplify the general term formula to make it easier to work with. Here, factoring out 15/2 shows a cleaner form. Always verify with n = 1 and n = 2 to check your formula.

(b) How many terms of the above G.P. will give the sum 945?
Answer: Using the sum formula S_n = a(r^n - 1)/(r - 1) with a = 15 and r = 2, we get 945 = 15(2^n - 1)/(2 - 1) = 15(2^n - 1). Simplifying, 2^n - 1 = 945/15 = 63, so 2^n = 64 = 2⁶. Therefore, n = 6. Hence, the sum of 6 terms of the G.P. is 945.
In simple words: Add the first 6 terms: 15 + 30 + 60 + 120 + 240 + 480 = 945.

Exam Tip: Express the final power as a power of the common ratio base (here, 2^n = 2⁶) to solve for n directly. Always check that 2^n equals a power of 2 before concluding n is an integer.

Question 23. The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal. Prove that: 2q = p + r, that is, p, q and r are in A.P.
Answer: Given that the equation (q - r)x² + (r - p)x + (p - q) = 0 has equal roots, the discriminant must be zero: b² - 4ac = 0. Here, a = (q - r), b = (r - p), c = (p - q). So (r - p)² - 4(q - r)(p - q) = 0. Expanding (r - p)²: r² + p² - 2pr. Expanding 4(q - r)(p - q): 4(qp - q² - rp + qr) = 4qp - 4q² - 4rp + 4qr. Combining: r² + p² - 2pr - 4qp + 4q² + 4rp - 4qr = 0. Simplifying: r² + p² + 2pr - 4qp - 4qr + 4q² = 0, which factors as (p + r)² - 4q(p + r) + 4q² = 0. Let y = p + r, then y² - 4qy + 4q² = 0, or (y - 2q)² = 0. So y = 2q, giving p + r = 2q. This proves that p, q, r are in A.P.
In simple words: If the quadratic has two equal roots, then p, q, r must form an arithmetic sequence where q is the middle term.

Exam Tip: When proving an A.P. relationship, always show that 2q = p + r (or equivalently, q - p = r - q). Discriminant = 0 is the key condition for equal roots. Factoring the resulting quadratic (y - 2q)² = 0 directly gives the result.