Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 10 Arithmetic Progression

Question 1. Which of the following sequences are in arithmetic progression?
(i) 2, 6, 10, 14,
(ii) 15, 12, 9, 6,
(iii) 5, 9, 12, 18,
(iv) \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots \)
Answer:
(i) 2, 6, 10, 14, ...
\( d_1 = 6 - 2 = 4 \)
\( d_2 = 10 - 6 = 4 \)
\( d_3 = 14 - 10 = 4 \)
Since \( d_1 = d_2 = d_3 \), the given sequence is in arithmetic progression.

(ii) 15, 12, 9, 6, ...
\( d_1 = 12 - 15 = -3 \)
\( d_2 = 9 - 12 = -3 \)
\( d_3 = 6 - 9 = -3 \)
Since \( d_1 = d_2 = d_3 \), the given sequence is in arithmetic progression.

(iii) 5, 9, 12, 18, ...
\( d_1 = 9 - 5 = 4 \)
\( d_2 = 12 - 9 = 3 \)
Since \( d_1 \neq d_2 \), the given sequence is not in arithmetic progression.

(iv) \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots \)
\( d_1 = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \)
\( d_2 = \frac{1}{4} - \frac{1}{3} = \frac{3 - 4}{12} = -\frac{1}{12} \)
Since \( d_1 \neq d_2 \), the given sequence is not in arithmetic progression.
In simple words: We check if the difference between any two next terms is always the same. If yes, it is A.P. If no, it is not A.P.

📝 Teacher's Note: Show students how to find differences step by step. Many students forget to check all pairs. Tell them to find at least 2-3 differences to be sure.

🎯 Exam Tip: Always write the formula first: "For A.P., common difference d must be same for all consecutive terms." Then show your calculations clearly.

Question 2. The nth term of sequence is (2n - 3), find its fifteenth term.
Answer:
nth term of A.P. = (2n - 3)
\( \Rightarrow t_n = 2n - 3 \)
\( \therefore 15^{th} \) term = \( t_{15} = 2 \times 15 - 3 = 30 - 3 = 27 \)
In simple words: We have a formula for any term. We put n = 15 to find the 15th term.

📝 Teacher's Note: Explain that we just substitute the value of n in the formula. This is like putting a number in place of n in a simple equation.

🎯 Exam Tip: Write "Given:" first, then substitute the value clearly. Show all steps even if they look simple.

Question 3. If the pth term of an A.P. is (2p + 3), find the A.P.
Answer:
pth term of an A.P. = 2p + 3
\( \Rightarrow t_p = 2p + 3 \)
Putting t = 1, 2, 3, ..., we get
\( t_1 = 2 \times 1 + 3 = 2 + 3 = 5 \)
\( t_2 = 2 \times 2 + 3 = 4 + 3 = 7 \)
\( t_3 = 2 \times 3 + 3 = 6 + 3 = 9 \) and so on.
Thus, the A.P. is 5, 7, 9, ...
In simple words: We find the first few terms by putting p = 1, 2, 3 in the formula. This gives us the actual sequence.

📝 Teacher's Note: Students often get confused between p and n. Explain that we replace p with 1, 2, 3 to get the first, second, third terms.

🎯 Exam Tip: Find at least the first 3 terms to show the pattern clearly. Write the A.P. in the standard form: a, a+d, a+2d, ...

Question 4. Find the 24th term of the sequence: 12, 10, 8, 6, ...
Answer:
The given sequence is 12, 10, 8, 6, ...
Now,
10 - 12 = -2
8 - 10 = -2
6 - 8 = -2, etc.
Hence, the given sequence is an A.P. with first term a = 12
and common difference d = -2.
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \Rightarrow t_{24} = 12 + (24 - 1)(-2) = 12 + 23 \times (-2) = 12 - 46 = -34 \)
So, the 24th term is -34.
In simple words: First we check if it is A.P. Then we use the formula to find the 24th term.

📝 Teacher's Note: Teach students to always verify that the sequence is A.P. before using the formula. Show them that negative common difference means the terms are decreasing.

🎯 Exam Tip: Write the formula \( t_n = a + (n-1)d \) clearly. Substitute values step by step. Check your answer - is it reasonable?

Question 5. Find the 30th term of the sequence: \( \frac{1}{2}, 1, \frac{3}{2}, \ldots \)
Answer:
The given sequence is \( \frac{1}{2}, 1, \frac{3}{2}, \ldots \)
Now,
\( 1 - \frac{1}{2} = \frac{1}{2} \)
\( \frac{3}{2} - 1 = \frac{1}{2} \), etc.
Hence, the given sequence is an A.P. with first term \( a = \frac{1}{2} \)
and common difference \( d = \frac{1}{2} \).
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \Rightarrow t_{30} = \frac{1}{2} + (30 - 1)\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{29}{2} = \frac{30}{2} = 15 \)
So, the 30th term is 15.
In simple words: We check differences, find they are all 1/2. Then use the formula to get the 30th term.

📝 Teacher's Note: When working with fractions, remind students to find common denominators carefully. Show them that adding many small fractions gives a big number.

🎯 Exam Tip: Write fractions clearly. When substituting in the formula, be careful with fraction arithmetic. Simplify your final answer.

Question 6. Find the 100th term of the sequence \( \sqrt{5}, 2\sqrt{5}, 3\sqrt{5}, \ldots \)
Answer:
The given A.P. is \( \sqrt{5}, 2\sqrt{5}, 3\sqrt{5}, \ldots \)
Now,
\( 2\sqrt{5} - \sqrt{5} = \sqrt{5} \)
\( 3\sqrt{5} - 2\sqrt{5} = \sqrt{5} \), etc.
Hence, the given sequence is an A.P. with first term \( a = \sqrt{5} \)
and common difference \( d = \sqrt{5} \).
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \Rightarrow t_{100} = \sqrt{5} + (100 - 1) \times \sqrt{5} = \sqrt{5} + 99\sqrt{5} = 100\sqrt{5} \)
So, the 100th term is \( 100\sqrt{5} \).
In simple words: Each term has √5 as a factor. The pattern is 1√5, 2√5, 3√5, so the 100th term is 100√5.

📝 Teacher's Note: Point out the pattern - the coefficient of √5 increases by 1 each time. This makes the calculation easier to see.

🎯 Exam Tip: When you see surds (√ symbols), look for the pattern in the coefficients. Keep the surd in your final answer unless asked to find decimal value.

Question 7. Find the 50th term of the sequence: \( \frac{1}{n}, \frac{n+1}{n}, \frac{2n+1}{n}, \ldots \)
Answer:
The given sequence is \( \frac{1}{n}, \frac{n+1}{n}, \frac{2n+1}{n}, \ldots \)
Now,
\( \frac{n+1}{n} - \frac{1}{n} = \frac{n+1-1}{n} = \frac{n}{n} = 1 \)
\( \frac{2n+1}{n} - \frac{n+1}{n} = \frac{2n+1-n-1}{n} = \frac{n}{n} = 1 \), etc.
Hence, the given sequence is an A.P. with first term \( a = \frac{1}{n} \)
and common difference d = 1.
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \Rightarrow t_{50} = \frac{1}{n} + (50 - 1)(1) = \frac{1}{n} + 49 \)
So, the 50th term is \( \frac{1}{n} + 49 \).
In simple words: Even though we have fractions with n, the difference between terms is always 1. So we add 49 to the first term.

📝 Teacher's Note: This question shows that A.P. can have variables (like n) in the terms. The key is that the common difference is still a constant.

🎯 Exam Tip: When finding differences with algebraic fractions, combine the fractions first. The common difference should be a number, not a fraction with variables.

Question 8. Is 402 a term of the sequence: 8, 13, 18, 23, ...?
Answer:
The given sequence is 8, 13, 18, 23, ...
Now,
13 - 8 = 5
18 - 13 = 5
23 - 18 = 5, etc.
Hence, the given sequence is an A.P. with first term a = 8
and common difference d = 5.
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \Rightarrow 402 = 8 + (n - 1)(5) \)
\( \Rightarrow 394 = 5n - 5 \)
\( \Rightarrow 399 = 5n \)
\( \Rightarrow n = \frac{399}{5} \)
The number of terms cannot be a fraction.
So clearly, 402 is not a term of the given sequence.
In simple words: We put 402 in the formula and solve for n. Since n is not a whole number, 402 is not in the sequence.

📝 Teacher's Note: Explain that n must always be a positive integer (1, 2, 3, ...). If we get a fraction or decimal, that number is not in the A.P.

🎯 Exam Tip: When checking if a number is in an A.P., substitute it as tn and solve for n. If n is not a positive integer, the number is not in the A.P.

Question 9. Find the common difference and 99th term of the arithmetic progression: \( 7\frac{3}{4}, 9\frac{1}{2}, 11\frac{1}{4}, \ldots \)
Answer:
The given A.P. is \( 7\frac{3}{4}, 9\frac{1}{2}, 11\frac{1}{4}, \ldots \)
i.e. \( \frac{31}{4}, \frac{19}{2}, \frac{45}{4}, \ldots \)
Common difference = \( d = \frac{19}{2} - \frac{31}{4} = \frac{38 - 31}{4} = \frac{7}{4} = 1\frac{3}{4} \)
First term = \( a = \frac{31}{4} \)
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \Rightarrow t_{99} = \frac{31}{4} + (99 - 1) \times \frac{7}{4} = \frac{31}{4} + 98 \times \frac{7}{4} = \frac{31 + 686}{4} = \frac{717}{4} = 179\frac{1}{4} \)
In simple words: We convert mixed numbers to improper fractions, find the common difference, then use the formula for the 99th term.

📝 Teacher's Note: Teach students to convert mixed numbers to improper fractions first. This makes calculations easier and avoids mistakes.

🎯 Exam Tip: When working with mixed numbers, convert to improper fractions immediately. Keep your working neat and check your fraction arithmetic carefully.

Question 10. How many terms are there in the series:
(i) 4, 7, 10, 13, ..., 148?
(ii) 0.5, 0.53, 0.56, ..., 1.1?
(iii) \( \frac{3}{4}, 1, 1\frac{1}{4}, \ldots, 3 \)
Answer: [Note: This question appears incomplete in the source, only showing the question text without the solution.]
In simple words: To find number of terms, we use the formula and solve for n when we know the last term.

📝 Teacher's Note: This is a reverse problem - we know the last term and need to find how many terms there are. Use tn = a + (n-1)d and solve for n.

🎯 Exam Tip: For "how many terms" questions, identify a, d, and the last term, then solve tn = a + (n-1)d for n. The answer should be a positive integer.

Solution:

(i) The given series is 4, 7, 10, 13, ...., 148
Answer:
7 - 4 = 3, 10 - 7 = 3, 13 - 10 = 3, etc
Thus, the given series is an A.P. with first term a = 4 and common difference d = 3.
Last term l = 148

Step 1: Use the A.P. formula
\( 4 + (n - 1)(3) = 148 \)
\( \implies (n - 1) \times 3 = 144 \)
\( \implies n - 1 = 48 \)
\( \implies n = 49 \)

Thus, there are 49 terms in the given series.
In simple words: We found the pattern (add 3 each time). Then we used the formula to count how many steps from 4 to 148.

📝 Teacher's Note: Show students how to find the common difference first. Check by subtracting any two consecutive terms. Then use the nth term formula.

🎯 Exam Tip: Always write "a = first term" and "d = common difference" clearly. Show the formula step by step to get full marks.

(ii) The given series is 0.5, 0.53, 0.56, ...., 1.1
Answer:
0.53 - 0.5 = 0.03, 0.56 - 0.53 = 0.03, etc.
Thus, the given series is an A.P. with first term a = 0.5 and common difference d = 0.03
Last term l = 1.1

Step 1: Use the A.P. formula
\( 0.5 + (n - 1)(0.03) = 1.1 \)
\( \implies (n - 1) \times 0.03 = 0.6 \)
\( \implies n - 1 = 20 \)
\( \implies n = 21 \)

Thus, there are 21 terms in the given series.
In simple words: Each number goes up by 0.03. We count how many steps from 0.5 to 1.1.

📝 Teacher's Note: With decimal A.P., students often make calculation errors. Tell them to be very careful with decimal subtraction.

🎯 Exam Tip: Write decimals clearly. Show the common difference calculation first. Double-check your arithmetic with decimals.

(iii) The given series is \( \frac{3}{4} \), 1, \( 1\frac{1}{4} \), ..., 3 ⇒ \( \frac{3}{4} \), 1, \( \frac{5}{4} \), ....., 3
Answer:
\( 1 - \frac{3}{4} = \frac{1}{4} \), \( \frac{5}{4} - 1 = \frac{1}{4} \), etc

Thus, the given series is an A.P. with first term \( a = \frac{3}{4} \) and common difference \( d = \frac{1}{4} \).

Last term l = 3

Step 1: Use the A.P. formula
\( \frac{3}{4} + (n - 1)\left(\frac{1}{4}\right) = 3 \)
\( \implies (n - 1) \times \frac{1}{4} = 3 - \frac{3}{4} \)
\( \implies (n - 1) \times \frac{1}{4} = \frac{9}{4} \)
\( \implies n - 1 = 9 \)
\( \implies n = 10 \)

Thus, there are 10 terms in the given series.
In simple words: Each fraction goes up by 1/4. We count from 3/4 to 3 in steps of 1/4.

📝 Teacher's Note: Convert mixed numbers to improper fractions first. Show students how to subtract fractions step by step.

🎯 Exam Tip: Always convert mixed numbers to improper fractions before calculating. Show all fraction arithmetic clearly.

Question 11. Which term of the A.P. 1 + 4 + 7 + 10 + ......... is 52?
Answer:
The given A.P. is 1 + 4 + 7 + 10 + ....
Here, first term a = 1 and common difference d = 4 - 1 = 3
Let nth term of the given A.P. be 52.

Step 1: Use the A.P. formula
\( \implies 52 = a + (n - 1)d \)
\( \implies 52 = 1 + (n - 1) \times 3 \)
\( \implies 51 = (n - 1) \times 3 \)
\( \implies n - 1 = 17 \)
\( \implies n = 18 \)

Thus, the 18th term of the given A.P. is 52.
In simple words: We know the pattern (add 3 each time) and want to find which position gives us 52.

📝 Teacher's Note: Emphasize that students must clearly identify a and d before substituting in the formula. Common mistake is using wrong d value.

🎯 Exam Tip: Write "Let nth term = 52" clearly. Show the substitution in the formula step by step. State your final answer clearly.

Question 12. If 5th and 6th terms of an A.P are respectively 6 and 5. Find the 11th term of the A.P
Answer:
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
Now, \( t_5 = 6 \)
\( \implies a + (5 - 1)d = 6 \)
\( \implies a + 4d = 6 \) ....(i)
And, \( t_6 = 5 \)
\( \implies a + (6 - 1)d = 5 \)
\( \implies a + 5d = 5 \) ....(ii)

Step 1: Subtract equation (i) from (ii)
\( -d = 1 \)
\( \implies d = -1 \)

Step 2: Substitute d = -1 in equation (i)
\( a + 4(-1) = 6 \)
\( \implies a - 4 = 6 \)
\( \implies a = 10 \)

Step 3: Find 11th term
\( \implies t_n = 10 + (n - 1)(-1) \)
\( \implies t_{11} = 10 + (11 - 1)(-1) = 10 - 10 = 0 \)

The 11th term is 0.
In simple words: We used two known terms to find the first term and common difference. Then we found the 11th term.

📝 Teacher's Note: Show students how to set up two equations from given information. Emphasize solving the system of equations step by step.

🎯 Exam Tip: Always write two equations clearly. Label them (i) and (ii). Show subtraction step clearly to find d first, then find a.

Question 13. If \( t_n \) represents nth term of an A.P, \( t_2 + t_5 - t_3 = 10 \) and \( t_2 + t_9 = 17 \), find its first term and its common difference
Answer:
Let the first term of an A.P. be a and the common difference be d.
The general term of an A.P. is given by \( t_n = a + (n - 1)d \)

Step 1: Use the first condition
Now, \( t_2 + t_5 - t_3 = 10 \)
\( \implies (a + d) + (a + 4d) - (a + 2d) = 10 \)
\( \implies a + d + a + 4d - a - 2d = 10 \)
\( \implies a + 3d = 10 \) ....(i)

Step 2: Use the second condition
Also, \( t_2 + t_9 = 17 \)
\( \implies (a + d) + (a + 8d) = 17 \)
\( \implies 2a + 9d = 17 \) ....(ii)

Step 3: Solve the system of equations
Multiplying equation (i) by 2, we get
\( 2a + 6d = 20 \) ....(iii)
Subtracting (ii) from (iii), we get
\( -3d = 3 \)
\( \implies d = -1 \)

Step 4: Find first term
Substituting value of d in (i), we get
\( a + 3(-1) = 10 \)
\( \implies a - 3 = 10 \)
\( \implies a = 13 \)

Hence, a = 13 and d = -1.
In simple words: We used the given conditions to make two equations with a and d. Then we solved them to find both values.

📝 Teacher's Note: Students often make sign errors when expanding brackets. Show each step of bracket expansion clearly on the board.

🎯 Exam Tip: Write each term formula clearly first. Show bracket expansion step by step. Label your equations and state final answer clearly.

Question 14. Find the 10th term from the end of the A.P. 4, 9, 14,......., 254
Answer:
The given A.P. is 4, 9, 14, ...., 254.
First term = 4
Common difference = 9 - 4 = 5
Last term l = 254

Step 1: Find the reverse A.P.
For the reverse A.P., first term = 254 and common difference = -5
Thus, 10th term from the end of given A.P.
= 10th term from the beginning of its reverse A.P.

Step 2: Calculate the term
\( = 254 + (10 - 1) \times (-5) \)
\( = 254 - 45 \)
\( = 209 \)

The 10th term from the end is 209.
In simple words: We reverse the A.P. by starting from 254 and going backwards. The 10th term from end becomes 10th term from start of the reversed series.

📝 Teacher's Note: Explain that "from the end" means we reverse the A.P. The common difference becomes negative when we go backwards.

🎯 Exam Tip: Always write "reverse A.P." clearly. Show that the common difference becomes negative. State your method before calculating.

Question 15. Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
Answer:
For an A.P.,
\( t_3 = 5 \)
\( \implies a + 2d = 5 \) ....(i)
And, \( t_7 = 9 \)
\( \implies a + 6d = 9 \) ....(ii)

Step 1: Find common difference
Subtracting (i) from (ii), we get
\( 4d = 4 \)
\( \implies d = 1 \)

Step 2: Find first term
Substituting d = 1 in (i), we get
\( a + 2 \times 1 = 5 \)
\( \implies a = 3 \)

Step 3: Write the A.P.
Thus, the required A.P. = a, a + d, a + 2d, a + 3d, ....
= 3, 4, 5, 6, ....

The arithmetic progression is 3, 4, 5, 6, .....
In simple words: We found the first term and common difference from the given information. Then we wrote the complete sequence.

📝 Teacher's Note: Show students that once we find a and d, we can write the complete A.P. by adding d repeatedly.

🎯 Exam Tip: Write at least the first 4-5 terms of the A.P. in your final answer. Show the pattern clearly.

Question 16. Find the 31st term of an A.P whose 10th term is 38 and 16th term is 74.
Answer:
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
Now, \( t_{10} = 38 \)
\( \implies a + 9d = 38 \) ....(i)
And, \( t_{16} = 74 \)
\( \implies a + 15d = 74 \) ....(ii)

Step 1: Find common difference
Subtracting (i) from (ii), we get
\( 6d = 36 \)
\( \implies d = 6 \)

Step 2: Find first term
Substituting d = 6 in (i), we get
\( a + 9 \times 6 = 38 \)
\( \implies a + 54 = 38 \)
\( \implies a = -16 \)

Step 3: Find 31st term
\( \implies t_n = -16 + (n - 1)(6) \)
\( \implies t_{31} = -16 + 30 \times 6 = -16 + 180 = 164 \)

The 31st term is 164.
In simple words: We used two known terms to find the pattern. Then we used the pattern to find the 31st term.

📝 Teacher's Note: Point out that the first term can be negative. Students often get confused when a is negative.

🎯 Exam Tip: Be careful with signs when a is negative. Show all arithmetic clearly. Double-check your final calculation.

Question 17. Which term of the series: 21, 18, 15, ............. is - 81? Can any term of this series be zero? If yes find the number of term.
Answer:
Part 1: Finding which term is -81
The given A.P. is 21, 18, 15, ....
Here, first term a = 21 and common difference d = 18 - 21 = -3
Let nth term of the given A.P. be - 81.

Step 1: Use A.P. formula
\( \implies -81 = a + (n - 1)d \)
\( \implies -81 = 21 + (n - 1) \times (-3) \)
\( \implies -102 = (n - 1) \times (-3) \)
\( \implies n - 1 = 34 \)
\( \implies n = 35 \)

Thus, the 35th term of the given A.P. is - 81.

Part 2: Checking if any term can be zero
Let pth term of this A.P. be 0.
\( \implies 21 + (p - 1) \times (-3) = 0 \)
\( \implies 21 - 3(p - 1) = 0 \)
\( \implies 21 - 3p + 3 = 0 \)
\( \implies 24 - 3p = 0 \)
\( \implies 3p = 24 \)
\( \implies p = 8 \)

Yes, the 8th term of this A.P. is 0.
In simple words: This sequence goes down by 3 each time. We found which position gives -81 and which position gives 0.

📝 Teacher's Note: Show students that the common difference is negative because the sequence is decreasing. Practice both types of questions.

🎯 Exam Tip: For decreasing A.P., d is negative. Answer both parts clearly. Show that you can find any specific term value.

Question 18. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31st term.
Answer:
Given:
Number of terms, n = 60
First term, a = 7
Last term, l = 125

Step 1: Find the common difference using last term formula.
\( t_{60} = 125 \)
\( \implies \) \( a + 59d = 125 \)
\( \implies \) \( 7 + 59d = 125 \)
\( \implies \) \( 59d = 118 \)
\( \implies \) \( d = 2 \)

Step 2: Find the 31st term.
\( t_{31} = a + 30d = 7 + 30(2) = 7 + 60 = 67 \)

The 31st term is 67
In simple words: We found how much the sequence goes up each time (common difference = 2). Then we used the formula to find the 31st number in the list.

📝 Teacher's Note: Show students that in A.P., we add the same number each time. Here we add 2 each time. The 31st term means we add 2 a total of 30 times to the first term.

🎯 Exam Tip: Always write "Given" first, then find d using the last term. Write the formula clearly: \( t_n = a + (n-1)d \). Show all steps to get full marks.

Question 19. The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34. Find the first three terms of the A.P.
Answer:
Let 'a' be the first term and 'd' be the common difference of the given A.P.

Step 1: Write equations from given conditions.
\( t_4 + t_8 = 24 \)
\( \implies \) \( (a + 3d) + (a + 7d) = 24 \)
\( \implies \) \( 2a + 10d = 24 \)
\( \implies \) \( a + 5d = 12 \) ....(i)

\( t_6 + t_{10} = 34 \)
\( \implies \) \( (a + 5d) + (a + 9d) = 34 \)
\( \implies \) \( 2a + 14d = 34 \)
\( \implies \) \( a + 7d = 17 \) ....(ii)

Step 2: Solve the equations.
Subtracting (i) from (ii):
\( 2d = 5 \)
\( \implies \) \( d = \frac{5}{2} = 2.5 \)

Step 3: Find first term.
From equation (i): \( a + 5(2.5) = 12 \)
\( \implies \) \( a + 12.5 = 12 \)
\( \implies \) \( a = -0.5 \)

Step 4: Find first three terms.
First term = a = -0.5
Second term = a + d = -0.5 + 2.5 = 2
Third term = a + 2d = -0.5 + 2(2.5) = 4.5

The first three terms are -0.5, 2, 4.5
In simple words: We made two equations using the given information. We solved them to find the first term and common difference. Then we found the first three numbers.

📝 Teacher's Note: Teach students to write \( t_n = a + (n-1)d \) first for each term. Make two simple equations. Solve by elimination method step by step.

🎯 Exam Tip: Always label equations as (i) and (ii). Show elimination clearly. Check your answer by putting values back in the original conditions.

Question 20. If the third term of an A.P. is 5 and the seventh terms is 9, find the 17th term.
Answer:
Let 'a' be the first term and 'd' be the common difference of the given A.P.

Step 1: Write equations from given information.
\( t_3 = 5 \)
\( \implies \) \( a + 2d = 5 \) ....(i)

\( t_7 = 9 \)
\( \implies \) \( a + 6d = 9 \) ....(ii)

Step 2: Solve for d.
Subtracting (i) from (ii):
\( 4d = 4 \)
\( \implies \) \( d = 1 \)

Step 3: Find first term.
From equation (i): \( a + 2(1) = 5 \)
\( \implies \) \( a = 3 \)

Step 4: Find 17th term.
\( t_{17} = a + 16d = 3 + 16(1) = 19 \)

The 17th term is 19
In simple words: We found that the sequence goes up by 1 each time (d = 1) and starts from 3. So the 17th term is 19.

📝 Teacher's Note: Show that we subtract to find d first. Once we know d, we can find a. This is the standard method for two-term problems.

🎯 Exam Tip: Write the term formulas clearly: \( t_3 = a + 2d \), \( t_7 = a + 6d \). Always find d first, then a, then the required term.

Exercise 10B

Question 1. In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Answer:
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)

Given:
\( 10 \times t_{10} = 30 \times t_{30} \)
\( \implies \) \( 10 \times (a + 9d) = 30 \times (a + 29d) \)
\( \implies \) \( a + 9d = 3 \times (a + 29d) \)
\( \implies \) \( a + 9d = 3a + 87d \)
\( \implies \) \( 2a + 78d = 0 \)
\( \implies \) \( a + 39d = 0 \)
\( \implies \) \( a = -39d \)

Now, \( t_{40} = a + 39d = -39d + 39d = 0 \)

The 40th term is 0
In simple words: We used the given condition to find a relationship between first term and common difference. This helped us find that the 40th term is zero.

📝 Teacher's Note: Explain that when we get \( a = -39d \), it means the first term and common difference have opposite signs. This makes the 40th term zero.

🎯 Exam Tip: Always simplify the given condition first. Write \( a \) in terms of \( d \) or vice versa. Then substitute in the required term formula.

Question 2. How many two-digit numbers are divisible by 3?
Answer:
The two-digit numbers divisible by 3 are as follows:
12, 15, 18, 21, ......, 99

Clearly, this forms an A.P. with first term, a = 12
and common difference, d = 3
Last term = nth term = 99

The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \implies \) \( 99 = 12 + (n - 1)(3) \)
\( \implies \) \( 99 = 12 + 3n - 3 \)
\( \implies \) \( 90 = 3n \)
\( \implies \) \( n = 30 \)

Thus, 30 two-digit numbers are divisible by 3
In simple words: We found all two-digit numbers that can be divided by 3. They form a pattern (A.P.) starting from 12 and ending at 99, with a difference of 3 each time.

📝 Teacher's Note: Show students that smallest two-digit multiple of 3 is 12 (since 9 is one-digit) and largest is 99. Count them using A.P. formula.

🎯 Exam Tip: For "divisible by n" problems, find first and last terms in the given range. Use A.P. formula to count total terms.

Question 3. Which term of A.P. 5, 15, 25 ........... will be 130 more than its 31st term?
Answer:
The given A.P. is 5, 15, 25, .....
Here, a = 5 and d = 15 - 5 = 10

Step 1: Find the 31st term.
\( t_{31} = a + 30d = 5 + 30 \times 10 = 5 + 300 = 305 \)

Step 2: Find which term is 130 more than 31st term.
Let the required term be nth term.
\( \therefore t_n - t_{31} = 130 \)
\( \implies \) \( [a + (n - 1)d] - 305 = 130 \)
\( \implies \) \( 5 + (n - 1)(10) = 435 \)
\( \implies \) \( (n - 1)(10) = 430 \)
\( \implies \) \( n - 1 = 43 \)
\( \implies \) \( n = 44 \)

Thus, required term = 44th term
In simple words: We found the 31st term first. Then we found which term is 130 more than that. It turned out to be the 44th term.

📝 Teacher's Note: Show that "130 more than 31st term" means we add 130 to the value of 31st term. Then find which term has this new value.

🎯 Exam Tip: First find the 31st term value. Then set up equation: \( t_n = t_{31} + 130 \). Solve for n to get the answer.

Question 4. Find the value of p, if x, 2x + p and 3x + 6 are in A.P
Answer:
Since x, 2x + p and 3x + 6 are in A.P., we have
\( (2x + p) - x = (3x + 6) - (2x + p) \)
\( \implies \) \( 2x + p - x = 3x + 6 - 2x - p \)
\( \implies \) \( x + p = x + 6 - p \)
\( \implies \) \( p + p = x - x + 6 \)
\( \implies \) \( 2p = 6 \)
\( \implies \) \( p = 3 \)

The value of p is 3
In simple words: In A.P., the difference between any two consecutive terms is the same. We used this rule to find p = 3.

📝 Teacher's Note: Teach that in A.P., \( b - a = c - b \) where a, b, c are consecutive terms. This is the key property to remember.

🎯 Exam Tip: Write the A.P. condition: "common difference is same". Set up equation: second term - first term = third term - second term.

Question 5. If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, which term of it is zero?
Answer:
For an A.P.,
\( t_3 = 4 \)
\( \implies \) \( a + 2d = 4 \) ....(i)
\( t_9 = -8 \)
\( \implies \) \( a + 8d = -8 \) ....(ii)

Subtracting (i) from (ii), we get
\( 6d = -12 \)
\( \implies \) \( d = -2 \)

Substituting d = -2 in (i), we get
\( a + 2(-2) = 4 \)
\( \implies \) \( a - 4 = 4 \)
\( \implies \) \( a = 8 \)
\( \implies \) General term = \( t_n = 8 + (n - 1)(-2) \)

Let pth term of this A.P. be 0.
\( \implies \) \( 8 + (p - 1) \times (-2) = 0 \)
\( \implies \) \( 8 - 2p + 2 = 0 \)
\( \implies \) \( 10 - 2p = 0 \)
\( \implies \) \( 2p = 10 \)
\( \implies \) \( p = 5 \)

Thus, 5th term of this A.P. is 0
In simple words: We found the first term and common difference. Then we found which term equals zero by setting the general term formula equal to 0.

📝 Teacher's Note: Show that common difference is negative, so terms are decreasing. Start from a = 8 and keep subtracting 2 to reach 0.

🎯 Exam Tip: Find a and d first. Then set \( t_n = 0 \) and solve for n. Always check: does the answer make sense with the sequence pattern?

Question 6. How many three-digit numbers are divisible by 87?
Answer:
The three-digit numbers divisible by 87 are as follows:
174, 261, ....., 957

Clearly, this forms an A.P. with first term, a = 174
and common difference, d = 87
Last term = nth term = 957

The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \implies \) \( 957 = 174 + (n - 1)(87) \)
\( \implies \) \( 783 = (n - 1) \times 87 \)
\( \implies \) \( 9 = n - 1 \)
\( \implies \) \( n = 10 \)

Thus, 10 three-digit numbers are divisible by 87
In simple words: We found the smallest and largest three-digit multiples of 87. Then counted how many such numbers exist using A.P. formula.

📝 Teacher's Note: Show students: smallest three-digit multiple of 87 is 2 × 87 = 174 (since 1 × 87 = 87 is two-digit). Largest is 11 × 87 = 957.

🎯 Exam Tip: Find first term by dividing 100 by 87 and taking next integer multiple. Find last term by dividing 999 by 87 and taking floor value multiple.

Question 7. For what value of n, the nth term of A.P 63, 65, 67, ........ and nth term of A.P. 3, 10, 17,........ are equal to each other?
Answer:
For A.P. 63, 65, 67, ........, we have a = 63 and d = 65 - 63 = 2
nth term = \( t_n = 63 + (n - 1) \times 2 \)

For A.P. 3, 10, 17, ........, we have a' = 3 and d' = 10 - 3 = 7
nth term = \( t_n' = 3 + (n - 1) \times 7 \)

The two A.P.s will have equal nth terms is
\( t_n = t_n' \)
\( \implies \) \( 63 + (n - 1) \times 2 = 3 + (n - 1) \times 7 \)
\( \implies \) \( 63 + 2n - 2 = 3 + 7n - 7 \)
\( \implies \) \( 61 + 2n = 7n - 4 \)
\( \implies \) \( 5n = 65 \)
\( \implies \) \( n = 13 \)

For n = 13, both A.P.s have equal nth terms
In simple words: We found the general terms of both sequences. Then we set them equal to find when they give the same value.

📝 Teacher's Note: Show students to write both general terms first. Then equate them and solve for n. Verify by calculating both 13th terms.

🎯 Exam Tip: Write both nth terms clearly. Set them equal and solve step by step. Always verify your answer by substituting back.

Question 8. Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Answer:
For given A.P.,
\( t_3 = a + 2d = 16 \) ....(i)
Now,
\( t_7 - t_5 = 12 \)
\( \implies (a + 6d) - (a + 4d) = 12 \)
\( \implies 2d = 12 \)
\( \implies d = 6 \)
Substituting the value of d in (i), we get
\( a + 2 \times 6 = 16 \)
\( \implies a + 12 = 16 \)
\( \implies a = 4 \)
Thus, the required A.P. = a, a + d, a + 2d, a + 3d, .....
= 4, 10, 16, 22, .....
In simple words: We found that the first term is 4 and each term increases by 6. So the sequence is 4, 10, 16, 22, and so on.

📝 Teacher's Note: Show students that A.P. means equal difference between consecutive terms. Here each term is 6 more than the previous term.

🎯 Exam Tip: Always write the general form first. Then substitute given values to find a and d. Write the final sequence clearly.

Question 9. If numbers n – 2, 4n – 1 and 5n + 2 are in A.P. find the value of n and its next two terms.
Answer:
Since (n – 2), (4n – 1) and (5n + 2) are in A.P., we have
(4n – 1) – (n – 2) = (5n + 2) – (4n – 1)
\( \implies 4n - 1 - n + 2 = 5n + 2 - 4n + 1 \)
\( \implies 3n + 1 = n + 3 \)
\( \implies 2n = 2 \)
\( \implies n = 1 \)
So, the given numbers are –1, 3, 7
\( \implies a = -1 \) and \( d = 3 - (-1) = 4 \)
Hence, the next two terms are (7 + 4) and (7 + 2 × 4)
i.e. 11 and 15.
In simple words: We found n = 1. So the three numbers become -1, 3, 7. The next two terms are 11 and 15.

📝 Teacher's Note: Explain that in A.P., the difference between any two consecutive terms is the same. Use this property to find n.

🎯 Exam Tip: Write the A.P. condition clearly: second term - first term = third term - second term. This gets you marks.

Question 10. Determine the value of k for which \( k^2 + 4k + 8 \), \( 2k^2 + 3k + 6 \) and \( 3k^2 + 4k + 4 \) are in A.P
Answer:
Since \( (k^2 + 4k + 8) \), \( (2k^2 + 3k + 6) \) and \( (3k^2 + 4k + 4) \) are in A.P., we have
\( (2k^2 + 3k + 6) - (k^2 + 4k + 8) = (3k^2 + 4k + 4) - (2k^2 + 3k + 6) \)
\( \implies 2k^2 + 3k + 6 - k^2 - 4k - 8 = 3k^2 + 4k + 4 - 2k^2 - 3k - 6 \)
\( \implies k^2 - k - 2 = k^2 + k - 2 \)
\( \implies 2k = 0 \)
\( \implies k = 0 \)
In simple words: We used the A.P. property that middle term minus first term equals third term minus middle term. This gave us k = 0.

📝 Teacher's Note: Show students how to expand and simplify algebraic expressions step by step. This is a common algebra mistake area.

🎯 Exam Tip: Write the A.P. condition first, then expand carefully. Check your algebra twice to avoid calculation mistakes.

Question 11. If a, b and c are in A.P show that:
(i) 4a, 4b and 4c are in A.P
(ii) a + 4, b + 4 and c + 4 are in A.P.
Answer:
a, b and c are in A.P.
\( \implies b - a = c - b \)
\( \implies 2b = a + c \)

(i) Given terms are 4a, 4b and 4c
Now, 4b – 4a = 2(2b – 2a)
= 2(a + c – 2a)
= 2(c – a)
And, 4c – 4b = 2(2c – 2b)
= 2(2c – a – c)
= 2(c – a)
Since 4b – 4a = 4c – 4b, the given terms are in A.P.

(ii) Given terms are (a + 4), (b + 4) and (c + 4)
Now, (b + 4) – (a + 4) = b – a
= \( \frac{a + c}{2} - a \)
= \( \frac{a + c - 2a}{2} \)
= \( \frac{c - a}{2} \)
And, (c + 4) – (b + 4) = c – b
= \( c - \frac{a + c}{2} \)
= \( \frac{2c - a - c}{2} \)
= \( \frac{c - a}{2} \)
Since (b + 4) – (a + 4) = (c + 4) – (b + 4), the given terms are in A.P.
In simple words: If three numbers are in A.P., then multiplying each by 4 or adding 4 to each keeps them in A.P.

📝 Teacher's Note: Use simple examples like 2, 4, 6 in A.P. Show that 8, 16, 24 and 6, 8, 10 are also in A.P.

🎯 Exam Tip: Always start with the given A.P. condition. Show that the common difference remains the same for the new terms.

Question 12. An A.P consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.
Answer:
Number of terms = n = 57
\( t_7 = 13 \)
\( \implies a + 6d = 13 \) ....(i)
Last term = \( t_{57} = 108 \)
\( \implies a + 56d = 108 \) ....(ii)
Subtracting (i) from (ii), we get
50d = 95
\( \implies d = \frac{95}{50} \)
\( \implies d = \frac{19}{10} \)
Substituting value of d in (i), we get
\( a + 6 \times \frac{19}{10} = 13 \)
\( \implies a + \frac{57}{5} = 13 \)
\( \implies a = 13 - \frac{57}{5} = \frac{65 - 57}{5} = \frac{8}{5} \)
\( \implies \) General term = \( t_n = \frac{8}{5} + (n - 1) \times \frac{19}{10} \)
\( \implies t_{45} = \frac{8}{5} + 44 \times \frac{19}{10} = \frac{8}{5} + \frac{418}{5} = \frac{426}{5} = 85.2 \)
In simple words: We found the first term and common difference using the 7th and 57th terms. Then we used the formula to find the 45th term.

📝 Teacher's Note: Teach students to set up two equations when given two terms of an A.P. This is a standard method.

🎯 Exam Tip: Always write the formula \( t_n = a + (n-1)d \) first. Show all working steps clearly for full marks.

Question 13. 4th term of an A.P is equal to 3 times its first term and 7th term exceeds twice the 3rd term by 1. Find the first term and the common difference.
Answer:
The general term of an A.P is given by \( t_n = a + (n - 1)d \)
Now, \( t_4 = 3 \times a \)
\( \implies a + 3d = 3a \)
\( \implies 2a - 3d = 0 \) ....(i)

Next, \( t_7 - 2 \times t_3 = 1 \)
\( \implies a + 6d - 2(a + 2d) = 1 \)
\( \implies a + 6d - 2a - 4d = 1 \)
\( \implies -a + 2d = 1 \) ....(ii)

Multiplying (ii) by 2, we get
-2a + 4d = 2 ....(iii)
Adding equations (i) and (iii), we get
d = 2
Substituting the value of d in (ii), we get
-a + 2 × 2 = 1
\( \implies -a + 4 = 1 \)
\( \implies a = 3 \)
Hence, a = 3 and d = 2
In simple words: We wrote two equations using the given conditions. Solving these equations, we found the first term is 3 and common difference is 2.

📝 Teacher's Note: Show students how to convert word problems into mathematical equations. Practice translating phrases like "exceeds by".

🎯 Exam Tip: Write two equations from the given conditions. Label them clearly as (i) and (ii). Show the solving steps.

Question 14. The sum of the 2nd term and the 7th term of an A.P is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P
Answer:
The general term of an A.P is given by \( t_n = a + (n - 1)d \)
Now, \( t_2 + t_7 = 30 \)
\( \implies (a + d) + (a + 6d) = 30 \)
\( \implies 2a + 7d = 30 \) ....(i)

Next, \( 2 \times t_8 - t_{15} = 1 \)
\( \implies 2 \times (a + 7d) - (a + 14d) = 1 \)
\( \implies 2a + 14d - a - 14d = 1 \)
\( \implies a = 1 \)
Substituting the value of a in (i), we get
2 × 1 + 7d = 30
\( \implies 7d = 28 \)
\( \implies d = 4 \)
Thus, required A.P. = a, a + d, a + 2d, a + 3d, .....
= 1, 5, 9, 13, .....
In simple words: We found that the first term is 1 and each term increases by 4. So the A.P. is 1, 5, 9, 13, and so on.

📝 Teacher's Note: Explain the meaning of "1 less than twice" - it means 2 × (something) - 1. Students often confuse this.

🎯 Exam Tip: Read the problem carefully. "15th term is 1 less than twice of 8th term" means \( t_{15} = 2t_8 - 1 \).

Question 15. In an A.P, if mth term is n and nth term is m, show that its rth term is (m + n – r)
Answer:
For an A.P.,
\( t_m = n \)
\( \Rightarrow a + (m - 1)d = n \) ....(i)
And, \( t_n = m \)
\( \Rightarrow a + (n - 1)d = m \) ....(ii)
Subtracting (i) from (ii), we get
\( (n - 1)d - (m - 1)d = m - n \)
\( \Rightarrow nd - d - md + d = m - n \)
\( \Rightarrow d(n - m) = m - n \)
\( \Rightarrow -d(m - n) = m - n \)
\( \Rightarrow d = -1 \)
Substituting \( d = -1 \) in equation (i), we get
\( a + (m - 1)(-1) = n \)
\( \Rightarrow a - m + 1 = n \)
\( \Rightarrow a = m + n - 1 \)

Now, \( t_r = a + (r - 1)d \)
= \( (m + n - 1) + (r - 1)(-1) \)
= \( m + n - 1 - r + 1 \)
= \( m + n - r \)
In simple words: We used the given conditions to find that the common difference is -1. Then we found the first term and used the formula to get the rth term.

📝 Teacher's Note: Show students that we get two equations from the given conditions. Solving these equations helps us find the pattern. The answer comes out neat when we subtract the equations.

🎯 Exam Tip: Write the two equations clearly from the given conditions. Show all algebra steps. The final answer must be exactly (m + n - r).

Question 16. Which term of the A.P 3, 10, 17, ………. Will be 84 more than its 13th term?
Answer:
The given A.P. is 3, 10, 17, …..
Here, \( a = 3 \) and \( d = 10 - 3 = 7 \)
Now,
\( t_{13} = a + 12d = 3 + 12 \times 7 = 3 + 84 = 87 \)
Let the required term be nth term.
\( \therefore t_n - t_{13} = 84 \)
\( \Rightarrow [a + (n - 1)d] - 87 = 84 \)
\( \Rightarrow 3 + (n - 1) \times 7 = 171 \)
\( \Rightarrow (n - 1) \times 7 = 168 \)
\( \Rightarrow n - 1 = 24 \)
\( \Rightarrow n = 25 \)
Thus, required term = 25th term
In simple words: We first found the 13th term. Then we set up an equation where some term is 84 more than the 13th term. Solving gives us the 25th term.

📝 Teacher's Note: Help students see that "84 more than" means we add 84 to the 13th term. This gives us the equation to solve for which term number.

🎯 Exam Tip: First find the 13th term clearly. Then write the equation "nth term = 13th term + 84". Show all steps to get n = 25.

Exercise 10 C

Question 1. Find the sum of the first 22 terms of the A.P.: 8, 3, -2, ……….
Answer:
The given A.P. is 8, 3, -2, …..
Here, \( a = 8, d = 3 - 8 = -5 \) and \( n = 22 \)
\( \therefore S = \frac{n}{2}[2a + (n - 1)d] \)
= \( \frac{22}{2}[2 \times 8 + (22 - 1) \times (-5)] \)
= \( 11[16 + 21 \times (-5)] \)
= \( 11[16 - 105] \)
= \( 11 \times (-89) \)
= \( -979 \)
In simple words: We used the sum formula for arithmetic progressions. Since the common difference is negative (-5), the sum becomes negative.

📝 Teacher's Note: Point out that the terms are getting smaller (8, 3, -2, -7...), so the sum will be negative. Students should expect this from the pattern.

🎯 Exam Tip: Write the formula first. Substitute values carefully. Since d = -5 is negative, be extra careful with signs in calculations.

Question 2. How many terms of the A.P. : 24, 21, 18, ……… must be taken so that their sum is 78?
Answer:
Let the number of terms taken be n.
The given A.P. is 24, 21, 18, …..
Here, \( a = 24 \) and \( d = 21 - 24 = -3 \)
\( S = \frac{n}{2}[2a + (n - 1)d] \)
\( \Rightarrow 78 = \frac{n}{2}[2 \times 24 + (n - 1) \times (-3)] \)
\( \Rightarrow 78 = \frac{n}{2}[48 - 3n + 3] \)
\( \Rightarrow 156 = n[51 - 3n] \)
\( \Rightarrow 156 = 51n - 3n^2 \)
\( \Rightarrow 3n^2 - 51n + 156 = 0 \)
\( \Rightarrow n^2 - 17n + 52 = 0 \)
\( \Rightarrow n^2 - 13n - 4n + 52 = 0 \)
\( \Rightarrow n(n - 13) - 4(n - 13) = 0 \)
\( \Rightarrow (n - 13)(n - 4) = 0 \)
\( \Rightarrow n = 13 \) or \( n = 4 \)
\( \therefore \) Required number of terms = 4 or 13
In simple words: We set up an equation where the sum of n terms equals 78. This gave us a quadratic equation with two solutions: 4 terms or 13 terms.

📝 Teacher's Note: Both answers are correct! With 4 terms, we get positive terms that sum to 78. With 13 terms, we include negative terms that also make the sum 78.

🎯 Exam Tip: Set up the sum equation carefully. When you get a quadratic, both solutions are usually valid. Check both answers if time permits.

Question 3. Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Answer:
nth term of an A.P. = \( t_n = 8n - 5 \)
Let a be the first term and d be the common difference of this A.P.
Then,
\( a = t_1 = 8 \times 1 - 5 = 8 - 5 = 3 \)
\( t_2 = 8 \times 2 - 5 = 16 - 5 = 11 \)
\( \therefore d = t_2 - t_1 = 11 - 3 = 8 \)

The sum of n terms of an A.P. = \( S = \frac{n}{2}[2a + (n - 1)d] \)
\( \Rightarrow \) Sum of 28 terms of an A.P. = \( \frac{28}{2}[2 \times 3 + 27 \times 8] \)
= \( 14[6 + 216] \)
= \( 14 \times 222 \)
= \( 3108 \)
In simple words: We used the nth term formula to find the first term and common difference. Then we used the sum formula to get 3108.

📝 Teacher's Note: Show students how to find 'a' and 'd' from the nth term formula. First term comes from putting n = 1. Common difference comes from subtracting consecutive terms.

🎯 Exam Tip: Find the first term by putting n = 1 in the given formula. Find d by calculating t₂ - t₁. Then use the standard sum formula.

Question 4(i). Find the sum of all odd natural numbers less than 50
Answer:
Odd natural numbers less than 50 are as follows:
1, 3, 5, 7, 9,………,49
Now, 3 - 1 = 2, 5 - 3 = 2 and so on.
Thus, this forms an A.P. with first term \( a = 1 \),
common difference \( d = 2 \) and last term \( l = 49 \)
Now, \( l = a + (n - 1)d \)
\( \Rightarrow 49 = 1 + (n - 1) \times 2 \)
\( \Rightarrow 48 = (n - 1) \times 2 \)
\( \Rightarrow 24 = n - 1 \)
\( \Rightarrow n = 25 \)

Sum of first n terms = \( S = \frac{n}{2}[a + l] \)
\( \Rightarrow \) Sum of odd natural numbers less than 50 = \( \frac{25}{2}[1 + 49] \)
= \( \frac{25}{2} \times 50 \)
= \( 25 \times 25 \)
= \( 625 \)
In simple words: All odd numbers form an A.P. with difference 2. We found there are 25 odd numbers less than 50, and their sum is 625.

📝 Teacher's Note: Help students see that odd numbers 1, 3, 5, 7... form an A.P. The pattern is easy to spot: add 2 each time.

🎯 Exam Tip: First write the sequence clearly. Find how many terms there are. Then use the sum formula with first and last terms.

Question 4(ii). Find the sum of first 12 natural numbers each of which is a multiple of 7.
Answer:
First 12 natural numbers which are multiple of 7 are as follows:
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84

Clearly, this forms an A.P. with first term \( a = 7 \),
common difference \( d = 7 \) and last term \( l = 84 \)

Sum of first n terms = \( S = \frac{n}{2}[a + l] \)
\( \Rightarrow \) Sum of first 12 natural numbers which are multiple of 7 = \( \frac{12}{2}[7 + 84] \)
= \( 6 \times 91 \)
= \( 546 \)
In simple words: Multiples of 7 form an A.P. with first term 7 and common difference 7. The 12th term is 84, and their sum is 546.

📝 Teacher's Note: Show students that multiples of any number form an A.P. The common difference is always that number itself.

🎯 Exam Tip: List the first few terms to see the pattern. For multiples of 7, the common difference is 7. Use the formula with first and last terms.

Question 5. Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Answer:
Given, \( t_2 = 14 \) and \( t_3 = 18 \)
\( \Rightarrow d = t_3 - t_2 = 18 - 14 = 4 \)
Now, \( t_2 = 14 \)
\( \Rightarrow a + d = 14 \)
\( \Rightarrow a + 4 = 14 \)
\( \Rightarrow a = 10 \)

Sum of n terms of an A.P. = \( \frac{n}{2}[2a + (n - 1)d] \)
\( \therefore \) Sum of first 51 terms of an A.P. = \( \frac{51}{2}[2 \times 10 + 50 \times 4] \)
= \( \frac{51}{2}[20 + 200] \)
= \( \frac{51}{2} \times 220 \)
= \( 51 \times 110 \)
= \( 5610 \)
In simple words: We found the common difference from the 2nd and 3rd terms. Then we found the first term. Using these in the sum formula gave us 5610.

📝 Teacher's Note: Teach students that consecutive terms have a difference equal to 'd'. So t₃ - t₂ = d. Then use t₂ to find 'a'.

🎯 Exam Tip: Find d first by subtracting consecutive terms. Then use any given term to find 'a'. Show all substitution steps clearly.

Question 6. The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms
Answer:
Sum of first 7 terms of an A.P = 49
\( \Rightarrow \frac{7}{2}[2a + 6d] = 49 \)
\( \Rightarrow \frac{7}{2} \times 2[a + 3d] = 49 \)
\( \Rightarrow 7[a + 3d] = 49 \)
\( \Rightarrow a + 3d = 7 \) ....(i)

Sum of first 17 terms of A.P. = 289
\( \Rightarrow \frac{17}{2}[2a + 16d] = 289 \)
\( \Rightarrow \frac{17}{2} \times 2[a + 8d] = 289 \)
\( \Rightarrow 17[a + 8d] = 289 \)
\( \Rightarrow a + 8d = 17 \) ....(ii)
In simple words: We got two equations from the given sums. Solving these equations will give us the values of 'a' and 'd' to find the general sum formula.

📝 Teacher's Note: Show students how to set up two equations from two given sums. This is a standard technique for finding 'a' and 'd' in A.P. problems.

🎯 Exam Tip: Write the sum formula and substitute the given values to get two equations. Solve these to find 'a' and 'd'. Then write the general sum formula.

Question 7. The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Answer:
Given:
First term a = 5
Last term l = 45
Sum of terms = 1000
Let there be n terms in this A.P.

Step 1: Use sum formula to find n.
Sum of first n terms = \( \frac{n}{2} [a + l] \)
\( \implies 1000 = \frac{n}{2} [5 + 45] \)
\( \implies 2000 = n \times 50 \)
\( \implies n = 40 \)

Step 2: Use last term formula to find d.
l = a + (n - 1)d
\( \implies 45 = 5 + (40 - 1)d \)
\( \implies 40 = 39d \)
\( \implies d = \frac{40}{39} \)

Number of terms = 40 and common difference = \( \frac{40}{39} \)
In simple words: We used the sum formula to find how many terms there are. Then we used the last term formula to find the common difference (how much we add each time).

📝 Teacher's Note: Show students that we always use the sum formula first to find n when we know first term, last term and sum. Then we find d using the nth term formula.

🎯 Exam Tip: Always write "Given" and list all values first. Show each step clearly. Write the final answer with proper labels for n and d.

Question 8. Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Answer:
Natural numbers between 250 and 1000 which are divisible by 9 are: 252, 261, 270, 279, ......., 999

Clearly, this forms an A.P. with first term a = 252, common difference d = 9 and last term l = 999

Step 1: Find number of terms.
l = a + (n - 1)d
\( \implies 999 = 252 + (n - 1) \times 9 \)
\( \implies 747 = (n - 1) \times 9 \)
\( \implies n - 1 = 83 \)
\( \implies n = 84 \)

Step 2: Find sum using sum formula.
Sum = \( \frac{n}{2} [a + l] = \frac{84}{2} [252 + 999] \)
= 42 × 1251
= 52542

Sum = 52542
In simple words: We first found which numbers between 250 and 1000 can be divided by 9. These form a pattern (A.P.). Then we counted how many such numbers exist and added them all up.

📝 Teacher's Note: Explain that multiples of 9 form an A.P. with common difference 9. Students should find the first and last terms carefully in the given range.

🎯 Exam Tip: Always identify the first term correctly (252, not 250). Write the A.P. clearly: a = 252, d = 9, l = 999. Show step-by-step calculation for full marks.

Question 9. The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Answer:
Let there be n terms in this A.P.
First term a = 34
Common difference d = 18
Last term l = 700

Step 1: Find number of terms.
l = a + (n - 1)d
\( \implies 700 = 34 + (n - 1) \times 18 \)
\( \implies (n - 1) \times 18 = 666 \)
\( \implies n - 1 = 37 \)
\( \implies n = 38 \)

Step 2: Find sum of terms.
Sum = \( \frac{n}{2} [a + l] = \frac{38}{2} [34 + 700] = 19 \times 734 = 13946 \)

Number of terms = 38 and Sum = 13946
In simple words: We used the formula for the last term to count how many terms are in this sequence. Then we used the sum formula to add all the terms together.

📝 Teacher's Note: Show students the two-step process: find n first using nth term formula, then find sum using sum formula. This is the standard approach.

🎯 Exam Tip: Write both formulas clearly: l = a + (n-1)d for finding n, and S = n/2[a+l] for sum. Double-check your arithmetic especially in multiplication.

Question 10. In an A.P, the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.
Answer:
First term a = 25
nth term = -17 \( \implies \) Last term l = -17
Sum of n terms = 132

Step 1: Find n using sum formula.
\( \frac{n}{2} [a + l] = 132 \)
\( \implies n(25 - 17) = 264 \)
\( \implies n \times 8 = 264 \)
\( \implies n = 33 \)

Step 2: Find common difference.
l = a + (n - 1)d
\( \implies -17 = 25 + 32d \)
\( \implies 32d = -42 \)
\( \implies d = -\frac{42}{32} = -\frac{21}{16} \)

n = 33 and d = \( -\frac{21}{16} \)
In simple words: We used the sum formula to find how many terms there are. The common difference is negative, which means each term gets smaller than the previous one.

📝 Teacher's Note: Point out that negative common difference means the sequence is decreasing. Show students how 25 becomes -17 by subtracting the same amount 32 times.

🎯 Exam Tip: When d is negative, write it as a proper fraction. Show that the sequence decreases when d < 0. Always simplify fractions to lowest terms.

Question 11. If the 8th term of an A.P is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
Answer:
For an A.P.
t₈ = 37
\( \implies a + 7d = 37 \) .....(i)
Also, t₁₅ - t₁₂ = 15
\( \implies (a + 14d) - (a + 11d) = 15 \)
\( \implies a + 14d - a - 11d = 15 \)
\( \implies 3d = 15 \)
\( \implies d = 5 \)

Substituting d = 5 in (i), we get
a + 7 × 5 = 37
\( \implies a + 35 = 37 \)
\( \implies a = 2 \)

Required A.P. = a, a + d, a + 2d, a + 3d, ....
= 2, 7, 12, 17, ......

Sum of first 20 terms = \( \frac{20}{2} [2 × 2 + 19 × 5] \)
= 10[4 + 95]
= 10 × 99
= 990

A.P. is 2, 7, 12, 17, ... and Sum of first 20 terms = 990
In simple words: We found the common difference first by using the given condition. Then we found the first term. The A.P. starts at 2 and increases by 5 each time.

📝 Teacher's Note: Show students how to set up two equations from the given conditions. Emphasize that t₁₅ - t₁₂ = 3d because terms are 3 positions apart.

🎯 Exam Tip: Always find d first when you have a condition like "one term exceeds another by some amount." Write the A.P. clearly as a sequence of numbers.

Question 12. Find the sum of all multiples of 7 between 300 and 700.
Answer:
Numbers between 300 and 700 which are multiples of 7 are: 301, 308, 315, 322, ....., 693

Clearly, this forms an A.P. with first term a = 301, common difference d = 7 and last term l = 693

Step 1: Find number of terms.
l = a + (n - 1)d
\( \implies 693 = 301 + (n - 1) × 7 \)
\( \implies 392 = (n - 1) × 7 \)
\( \implies n - 1 = 56 \)
\( \implies n = 57 \)

Step 2: Find sum.
Sum = \( \frac{n}{2} [a + l] = \frac{57}{2} [301 + 693] \)
= \( \frac{57}{2} × 994 \)
= 57 × 497
= 28329

Sum = 28329
In simple words: We found all numbers between 300 and 700 that can be divided exactly by 7. These form a pattern and we added them all up.

📝 Teacher's Note: Help students identify that 301 is the first multiple of 7 after 300 (since 7 × 43 = 301). Common mistake is starting with 300 itself.

🎯 Exam Tip: Find the correct first term by dividing 300 by 7 and taking the next whole multiple. Always verify that both first and last terms are within the given range.

Question 13. The sum of n natural numbers is 5n² + 4n. Find its 8th term
Answer:
Sum of n natural numbers = Sₙ = 5n² + 4n
\( \implies \) Sum of (n - 1) natural numbers = Sₙ₋₁ = 5(n - 1)² + 4(n - 1)
= 5(n² + 1 - 2n) + 4n - 4
= 5n² + 5 - 10n + 4n - 4
= 5n² - 6n + 1

nth term = Sₙ - Sₙ₋₁ = 5n² + 4n - 5n² + 6n - 1 = 10n - 1
\( \implies \) 8th term = t₈ = 10 × 8 - 1 = 80 - 1 = 79

8th term = 79
In simple words: We found a formula for any term by subtracting the sum of (n-1) terms from the sum of n terms. Then we put n = 8 to get the 8th term.

📝 Teacher's Note: Teach students the method: nth term = Sₙ - Sₙ₋₁. This is a standard technique when sum formula is given instead of the sequence itself.

🎯 Exam Tip: Write the general formula for nth term first (10n - 1), then substitute the required value. Show that this gives a valid A.P. with first term 9 and common difference 10.

Question 14. The fourth term of an A.P. is 11 and the term exceeds twice the fourth term by 5 the A.P and the sum of first 50 terms
Answer:
For an A.P.
t₄ = 11
\( \implies a + 3d = 11 \) .....(i)
Also, t₈ - 2t₄ = 5
\( \implies (a + 7d) - 2 × 11 = 5 \)
\( \implies a + 7d - 22 = 5 \)
\( \implies a + 7d = 27 \) .....(ii)

Subtracting (i) from (ii), we get
4d = 16
\( \implies d = 4 \)

Substituting d = 4 in (i), we get
a + 3 × 4 = 11
\( \implies a + 12 = 11 \)
\( \implies a = -1 \)

Required A.P. = a, a + d, a + 2d, a + 3d, ....
= -1, 3, 7, 11, ......

Sum of first 50 terms = \( \frac{50}{2} [2 × (-1) + 49 × 4] \)
= 25[-2 + 196]
= 25 × 194
= 4850

A.P. is -1, 3, 7, 11, ... and Sum of first 50 terms = 4850
In simple words: We used the given conditions to find the first term and common difference. The sequence starts at -1 and increases by 4 each time.

📝 Teacher's Note: Help students understand that "exceeds twice the fourth term by 5" means t₈ = 2t₄ + 5. Set up equations carefully from word problems.

🎯 Exam Tip: Read the question carefully to identify which term "exceeds" which other term. Write both conditions as separate equations and solve systematically.

Exercise 10 D

Question 1. Find three numbers in A.P. whose sum is 24 and whose product is 440.
Answer:
Let the three numbers in A.P. be \( a - d \), \( a \) and \( a + d \).

Step 1: Find the middle term using sum condition.
\( (a - d) + a + (a + d) = 24 \)
\( \implies 3a = 24 \)
\( \implies a = 8 \) ....(i)

Step 2: Use product condition to find d.
Also, \( (a - d) \times a \times (a + d) = 440 \)
\( \implies (a^2 - d^2) \times a = 440 \)
\( \implies (8^2 - d^2) \times 8 = 440 \) .....[From (i)]
\( \implies 64 - d^2 = 55 \)
\( \implies d^2 = 9 \)
\( \implies d = \pm 3 \)

Step 3: Find the required terms.
When \( a = 8 \) and \( d = 3 \):
Required terms = \( a - d \), \( a \) and \( a + d \)
= \( 8 - 3 \), \( 8 \), \( 8 + 3 \)
= 5, 8, 11

When \( a = 8 \) and \( d = -3 \):
Required terms = \( a - d \), \( a \) and \( a + d \)
= \( 8 - (-3) \), \( 8 \), \( 8 + (-3) \)
= 11, 8, 5

The three numbers are 5, 8, 11 (or 11, 8, 5).
In simple words: We use the fact that in A.P., the middle term is the average of the three numbers. Then we use the product condition to find the common difference.

📝 Teacher's Note: Show students that both answers (5, 8, 11) and (11, 8, 5) are correct. They are the same A.P. written in different order. Always check by substituting back.

🎯 Exam Tip: Always write "Let the three numbers be a-d, a, a+d" at the start. This standard form makes the algebra much easier. Don't forget to check both values of d.

Question 2. The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.
Answer:
Let the three consecutive terms in A.P. be \( a - d \), \( a \) and \( a + d \).

Step 1: Find the middle term using sum condition.
\( (a - d) + a + (a + d) = 21 \)
\( \implies 3a = 21 \)
\( \implies a = 7 \) ....(i)

Step 2: Use sum of squares condition.
Also, \( (a - d)^2 + a^2 + (a + d)^2 = 165 \)
\( \implies a^2 + d^2 - 2ad + a^2 + a^2 + d^2 + 2ad = 165 \)
\( \implies 3a^2 + 2d^2 = 165 \)
\( \implies 3 \times (7)^2 + 2d^2 = 165 \) .....[From (i)]
\( \implies 3 \times 49 + 2d^2 = 165 \)
\( \implies 147 + 2d^2 = 165 \)
\( \implies 2d^2 = 18 \)
\( \implies d^2 = 9 \)
\( \implies d = \pm 3 \)

Step 3: Find the required terms.
When \( a = 7 \) and \( d = 3 \):
Required terms = \( a - d \), \( a \) and \( a + d \)
= \( 7 - 3 \), \( 7 \), \( 7 + 3 \)
= 4, 7, 10

When \( a = 7 \) and \( d = -3 \):
Required terms = \( a - d \), \( a \) and \( a + d \)
= \( 7 - (-3) \), \( 7 \), \( 7 + (-3) \)
= 10, 7, 4

The three terms are 4, 7, 10 (or 10, 7, 4).
In simple words: We first find the middle number from the sum. Then we use the fact that squaring removes minus signs, so we get an equation to find the common difference.

📝 Teacher's Note: When expanding \( (a-d)^2 \) and \( (a+d)^2 \), the cross terms cancel out neatly. Show this step clearly so students understand why we get \( 3a^2 + 2d^2 \).

🎯 Exam Tip: Always expand squares carefully. Write \( (a-d)^2 = a^2 - 2ad + d^2 \) clearly. Students often make sign errors here. Check your answer by substituting back.

Question 3. The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Answer:
Let the four angles of a quadrilateral in A.P. be \( a \), \( a + 20° \), \( a + 40° \) and \( a + 60° \).

Step 1: Use angle sum property of quadrilateral.
\( a + (a + 20°) + (a + 40°) + (a + 60°) = 360° \) .....[Angle sum property]
\( \implies 4a + 120° = 360° \)
\( \implies 4a = 240° \)
\( \implies a = 60° \) ....(i)

Step 2: Find all angles.
Thus, angles of a quadrilateral are \( a \), \( a + 20° \), \( a + 40° \) and \( a + 60° \)
= \( 60° \), \( 80° \), \( 100° \) and \( 120° \)

The angles are 60°, 80°, 100° and 120°.
In simple words: The four angles of any quadrilateral always add up to 360°. We use this fact along with the A.P. pattern to find the first angle, then calculate the rest.

📝 Teacher's Note: Remind students that the sum of interior angles of any quadrilateral is always 360°. Draw a quadrilateral and show how it can be divided into two triangles, each with angle sum 180°.

🎯 Exam Tip: Write "Angle sum property of quadrilateral" when using the 360° fact. This shows the examiner you know the property. Always check that your four angles add to 360°.

Question 4. Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.
Answer:
Let the four parts be \( (a - 3d) \), \( (a - d) \), \( (a + d) \) and \( (a + 3d) \).

Step 1: Use sum condition.
Then, \( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 96 \)
\( \implies 4a = 96 \)
\( \implies a = 24 \)

Step 2: Use ratio of products condition.
It is given that
\( \frac{(a - d) \times (a + d)}{(a - 3d) \times (a + 3d)} = \frac{15}{7} \)

\( \implies \frac{a^2 - d^2}{a^2 - 9d^2} = \frac{15}{7} \)

\( \implies \frac{576 - d^2}{576 - 9d^2} = \frac{15}{7} \)

\( \implies 4032 - 7d^2 = 8640 - 135d^2 \)
\( \implies 128d^2 = 4608 \)
\( \implies d^2 = 36 \)
\( \implies d = \pm 6 \)

Step 3: Find the four parts.
When \( a = 24 \), \( d = 6 \):
\( a - 3d = 24 - 3(6) = 6 \)
\( a - d = 24 - 6 = 18 \)
\( a + d = 24 + 6 = 30 \)
\( a + 3d = 24 + 3(6) = 42 \)

When \( a = 24 \), \( d = -6 \):
\( a - 3d = 24 - 3(-6) = 42 \)
\( a - d = 24 - (-6) = 30 \)
\( a + d = 24 + (-6) = 18 \)
\( a + 3d = 24 + 3(-6) = 6 \)

Thus, the four parts are (6, 18, 30, 42) or (42, 30, 18, 6).
In simple words: Means are the middle two terms, extremes are the first and last terms. We use the given ratio of their products along with the sum condition to find the parts.

📝 Teacher's Note: Explain that means = middle terms and extremes = end terms. For four terms in A.P., use the pattern a-3d, a-d, a+d, a+3d to keep equal spacing.

🎯 Exam Tip: When writing ratio of products, identify means and extremes first. Means = 2nd and 3rd terms, Extremes = 1st and 4th terms. Cross multiply carefully when solving the ratio.

Question 5. Find five numbers in A.P. whose sum is \( 12\frac{1}{2} \) and the ratio of the first to the last terms is 2: 3.
Answer:
Let the five numbers in A.P. be \( (a - 2d) \), \( (a - d) \), \( a \), \( (a + d) \) and \( (a + 2d) \).

Step 1: Use sum condition.
Then, \( (a - 2d) + (a - d) + a + (a + d) + (a + 2d) = 12\frac{1}{2} \)

\( \implies 5a = \frac{25}{2} \)

\( \implies a = \frac{5}{2} \)

Step 2: Use ratio condition.
It is given that
\( \frac{a - 2d}{a + 2d} = \frac{2}{3} \)
\( \implies 3a - 6d = 2a + 4d \)
\( \implies a = 10d \)
\( \implies \frac{5}{2} = 10d \)

\( \implies d = \frac{1}{4} \)

Step 3: Find all five numbers.
\( \implies a = \frac{5}{2} \) and \( d = \frac{1}{4} \)

Thus, we have
\( a - 2d = \frac{5}{2} - 2 \times \frac{1}{4} = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2 \)

\( a - d = \frac{5}{2} - \frac{1}{4} = \frac{10-1}{4} = \frac{9}{4} \)

\( a = \frac{5}{2} \)

\( a + d = \frac{5}{2} + \frac{1}{4} = \frac{10+1}{4} = \frac{11}{4} \)

\( a + 3d = \frac{5}{2} + 2 \times \frac{1}{4} = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3 \)

Thus, the five numbers in A.P. are \( 2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4} \) and 3
= 2, 2.25, 2.5, 2.75 and 3
In simple words: We use the sum to find the middle term, then use the ratio of first to last term to find the common difference. For five terms, use the pattern a-2d, a-d, a, a+d, a+2d.

📝 Teacher's Note: For five terms in A.P., the middle term is the average of all terms. Show students that the sum of five terms = 5 × middle term. Convert fractions to decimals to help check answers.

🎯 Exam Tip: Write fractions clearly with proper fraction notation. When given mixed fractions like \( 12\frac{1}{2} \), convert to improper fraction \( \frac{25}{2} \) first. Always verify your ratio condition at the end.

Question 6. Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
Answer:
Let the three parts in A.P. be \( (a - d) \), \( a \) and \( (a + d) \).

Step 1: Use sum condition.
Then, \( (a - d) + a + (a + d) = 207 \)
\( \implies 3a = 207 \)
\( \implies a = 69 \)

Step 2: Use product condition.
It is given that
\( (a - d) \times a = 4623 \)
\( \implies (69 - d) \times 69 = 4623 \)
\( \implies 69 - d = 67 \)
\( \implies d = 2 \)
\( \implies a = 69 \) and \( d = 2 \)

Step 3: Find the three parts.
Thus, we have
\( a - d = 69 - 2 = 67 \)
\( a = 69 \)
\( a + d = 69 + 2 = 71 \)

Thus, the three parts in A.P are 67, 69 and 71.
In simple words: We find the middle term from the sum condition. Then we use the fact that the two smaller parts are the first and middle terms to find the common difference.

📝 Teacher's Note: Point out that "two smaller parts" means the smallest and middle parts, not the two largest parts. Students sometimes confuse this and use wrong terms in the product.

🎯 Exam Tip: Identify which are the "two smaller parts" correctly. In this case, it's the 1st and 2nd terms, not 1st and 3rd. Always check: 67 + 69 + 71 = 207 ✓ and 67 × 69 = 4623 ✓

Question 7. The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers.
Answer:
Let the three numbers in A.P. be \( (a - d), a \) and \( (a + d) \).
Then, \( (a - d) + a + (a + d) = 15 \)
\( \Rightarrow 3a = 15 \)
\( \Rightarrow a = 5 \)
It is given that
\( (a - d)^2 + (a + d)^2 = 58 \)
\( \Rightarrow a^2 + d^2 - 2ad + a^2 + d^2 + 2ad = 58 \)
\( \Rightarrow 2a^2 + 2d^2 = 58 \)
\( \Rightarrow 2(a^2 + d^2) = 58 \)
\( \Rightarrow a^2 + d^2 = 29 \)
\( \Rightarrow 5^2 + d^2 = 29 \)
\( \Rightarrow 25 + d^2 = 29 \)
\( \Rightarrow d^2 = 4 \)
\( \Rightarrow d = \pm 2 \)

When \( a = 5 \) and \( d = 2 \):
\( a - d = 5 - 2 = 3 \)
\( a = 5 \)
\( a + d = 5 + 2 = 7 \)

When \( a = 5 \) and \( d = -2 \):
\( a - d = 5 - (-2) = 7 \)
\( a = 5 \)
\( a + d = 5 + (-2) = 3 \)

Thus, the three numbers in A.P are \( (3, 5, 7) \) or \( (7, 5, 3) \).
In simple words: We used the standard form for three numbers in A.P. We found the middle term first, then the common difference. Both answers give the same three numbers in different order.

📝 Teacher's Note: Always start by writing three numbers as (a-d), a, (a+d). This makes the sum very easy to calculate. Show students that both solutions are the same numbers in reverse order.

🎯 Exam Tip: Write both possible values of d and find both sets of numbers. The examiner wants to see that you checked both positive and negative values of the common difference.

Question 8. Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Answer:
Let the four numbers in A.P. be \( (a - 3d), (a - d), (a + d) \) and \( (a + 3d) \).
Then, \( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20 \)
\( \Rightarrow 4a = 20 \)
\( \Rightarrow a = 5 \)
It is given that
\( (a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 120 \)
\( \Rightarrow a^2 + 9d^2 - 6ad + a^2 + d^2 - 2ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad = 120 \)
\( \Rightarrow 4a^2 + 20d^2 = 120 \)
\( \Rightarrow a^2 + 5d^2 = 30 \)
\( \Rightarrow 5^2 + 5d^2 = 30 \)
\( \Rightarrow 25 + 5d^2 = 30 \)
\( \Rightarrow 5d^2 = 5 \)
\( \Rightarrow d^2 = 1 \)
\( \Rightarrow d = \pm 1 \)

When \( a = 5, d = 1 \):
\( a - 3d = 5 - 3(1) = 2 \)
\( a - d = 5 - 1 = 4 \)
\( a + d = 5 + 1 = 6 \)
\( a + 3d = 5 + 3(1) = 8 \)

When \( a = 5, d = -1 \):
\( a - 3d = 5 - 3(-1) = 8 \)
\( a - d = 5 - (-1) = 6 \)
\( a + d = 5 + (-1) = 4 \)
\( a + 3d = 5 + 3(-1) = 2 \)

Thus, the four parts are \( (2, 4, 6, 8) \) or \( (8, 6, 4, 2) \).
In simple words: We take four numbers equally spaced around the middle value. The spacing is the common difference. Both answers are the same numbers in reverse order.

📝 Teacher's Note: For four numbers in A.P., use the form (a-3d), (a-d), (a+d), (a+3d). This keeps the calculations symmetric and neat. The coefficients are always 3, 1, 1, 3.

🎯 Exam Tip: Check your answer by adding the numbers to get 20 and adding their squares to get 120. This verification shows the examiner you understand the problem completely.

Question 9. Insert one arithmetic mean between 3 and 13.
Answer:
Arithmetic mean between 3 and 13 = \( \frac{3 + 13}{2} = \frac{16}{2} = 8 \)
In simple words: To find the number exactly in the middle of two numbers, we add them and divide by 2.

📝 Teacher's Note: Show students that this is like finding the average of two numbers. The arithmetic mean is always exactly in the middle, making three numbers in A.P.

🎯 Exam Tip: Use the formula (a+b)/2 for arithmetic mean between two numbers. Always check: 3, 8, 13 should have equal differences (5 each).

Question 10. The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Answer:
Let the number of sides of a polygon be n.
The smallest angle = 120° = a
Common difference in angles = d = 5°
Now, in a polygon of n sides, the sum of interior angles = \( (2n - 4) \times 90° \)
\( \Rightarrow \frac{n}{2}[2 \times 120° + (n - 1) \times 5°] = (2n - 4) \times 90° \)
\( \Rightarrow \frac{n}{2}[240° + 5n - 5°] = 180n - 360° \)
\( \Rightarrow n[235° + 5n] = 360n - 720° \)
\( \Rightarrow 235n + 5n^2 = 360n - 720 \)
\( \Rightarrow 5n^2 - 125n + 720 = 0 \)
\( \Rightarrow n^2 - 25n + 144 = 0 \)
\( \Rightarrow n^2 - 16n - 9n + 144 = 0 \)
\( \Rightarrow n(n - 16) - 9(n - 16) = 0 \)
\( \Rightarrow (n - 16)(n - 9) = 0 \)
\( \Rightarrow n = 16 \) or \( n = 9 \)
In simple words: We used the fact that all interior angles of any polygon must add up to a fixed total. Then we solved to find how many sides the polygon has.

📝 Teacher's Note: Remind students that the sum of interior angles of an n-sided polygon is (2n-4)×90°. This is a key formula. Check which answer makes sense by seeing if all angles are reasonable.

🎯 Exam Tip: Write the sum formula clearly first. Check both answers: for n=9, largest angle = 120°+8×5° = 160° (good). For n=16, largest angle = 120°+15×5° = 195° (too big for interior angle).

Question 11. \( \frac{1}{a}, \frac{1}{b} \) and \( \frac{1}{c} \) are in A.P. Show that: bc, ca and ab are also in A.P.
Answer:
\( \frac{1}{a}, \frac{1}{b} \) and \( \frac{1}{c} \) are in A.P.
\( \Rightarrow \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \)
\( \Rightarrow \frac{a - b}{ab} = \frac{b - c}{bc} \)
\( \Rightarrow \frac{a - b}{a} = \frac{b - c}{c} \)
\( \Rightarrow ac - bc = ab - ac \)
\( \Rightarrow ac + ac = ab + bc \)
\( \Rightarrow 2ac = ab + bc \)
\( \Rightarrow 2ca = ab + bc \)
\( \Rightarrow \) bc, ca and ab are also in A.P.
In simple words: We used the definition of A.P. (equal differences) to show that if the reciprocals are in A.P., then the products are also in A.P.

📝 Teacher's Note: Start with the definition that in A.P., the middle term minus first term equals third term minus middle term. Cross-multiply carefully and rearrange to get the required form.

🎯 Exam Tip: Write the A.P. condition clearly as equal differences first. Show all algebraic steps. End by writing "Hence, bc, ca, ab are in A.P." to complete the proof.

Question 12. Insert four A.M.s between 14 and -1.
Answer:
Let the required arithmetic means (A.M.s) between 14 and -1 be \( A_1, A_2, A_3 \) and \( A_4 \).
\( \Rightarrow 14, A_1, A_2, A_3, A_4 \) and -1 are in A.P.
\( \Rightarrow 14 = \) First term
\( \Rightarrow -1 = 6^{th} \) term of this A.P.
\( \Rightarrow -1 = 14 + 5d \)
\( \Rightarrow 5d = -15 \)
\( \Rightarrow d = -3 \)

\( \Rightarrow A_1 = 14 + d = 14 + (-3) = 11 \)
\( A_2 = 14 + 2d = 14 + 2(-3) = 8 \)
\( A_3 = 14 + 3d = 14 + 3(-3) = 5 \)
\( A_4 = 14 + 4d = 14 + 4(-3) = 2 \)

Hence, required A.M.s between 14 and -1 = 11, 8, 5 and 2
In simple words: We made an A.P. with 6 terms where first is 14 and last is -1. Then we found the 4 middle terms using the common difference.

📝 Teacher's Note: Show students that inserting n arithmetic means between two numbers creates an A.P. with (n+2) terms total. The 6th term formula is a + 5d.

🎯 Exam Tip: Always identify which term the last number is (here -1 is the 6th term). Use the formula a_n = a + (n-1)d to find the common difference first.

Question 13. Insert five A.M.s between -12 and 8.
Answer:
Let the required arithmetic means (A.M.s) between -12 and 8 be \( A_1, A_2, A_3, A_4 \) and \( A_5 \).
\( \Rightarrow -12, A_1, A_2, A_3, A_4, A_5 \) and 8 are in A.P.
\( \Rightarrow -12 = \) First term
\( \Rightarrow 8 = 7^{th} \) term of this A.P.
\( \Rightarrow 8 = -12 + 6d \)
\( \Rightarrow 6d = 20 \)
\( \Rightarrow d = \frac{10}{3} \)

\( \Rightarrow A_1 = -12 + d = -12 + \frac{10}{3} = \frac{-36 + 10}{3} = \frac{-26}{3} \)
\( A_2 = -12 + 2d = -12 + \frac{20}{3} = \frac{-36 + 20}{3} = \frac{-16}{3} \)
\( A_3 = -12 + 3d = -12 + \frac{30}{3} = \frac{-36 + 30}{3} = \frac{-6}{3} \)
\( A_4 = -12 + 4d = -12 + \frac{40}{3} = \frac{-36 + 40}{3} = \frac{-4}{3} \)
\( A_5 = -12 + 5d = -12 + \frac{50}{3} = \frac{-36 + 50}{3} = \frac{-14}{3} \)

Hence, required A.M.s between -12 and 8 = \( \frac{-26}{3}, \frac{-16}{3}, \frac{-6}{3}, \frac{-4}{3} \) and \( \frac{-14}{3} \)
In simple words: We divided the gap between -12 and 8 into 6 equal parts. Each part has length 10/3. We added this step by step to get the 5 middle numbers.

📝 Teacher's Note: When the common difference is a fraction, be extra careful with calculations. Show students how to add fractions by making common denominators.

🎯 Exam Tip: Write fractions in simplest form. Check your answer: the 7th term should be 8 when you substitute. Always keep fractions unless asked for decimals.

Question 14. Insert six A.M.s between 15 and -15.
Answer:
Let the required arithmetic means (A.M.s) between 15 and -15 be \( A_1, A_2, A_3, A_4, A_5 \) and \( A_6 \).
\( \Rightarrow 15, A_1, A_2, A_3, A_4, A_5, A_6 \) and -15 are in A.P.
\( \Rightarrow 15 = \) First term
\( \Rightarrow -15 = 8^{th} \) term of this A.P.
\( \Rightarrow -15 = 15 + 7d \)
\( \Rightarrow 7d = -30 \)
\( \Rightarrow d = \frac{-30}{7} \)

\( \Rightarrow A_1 = 15 + d = 15 + \frac{-30}{7} = \frac{105 - 30}{7} = \frac{75}{7} \)
\( A_2 = 15 + 2d = 15 + \frac{-60}{7} = \frac{105 - 60}{7} = \frac{45}{7} \)
\( A_3 = 15 + 3d = 15 + \frac{-90}{7} = \frac{105 - 90}{7} = \frac{15}{7} \)
\( A_4 = 15 + 4d = 15 + \frac{-120}{7} = \frac{105 - 120}{7} = \frac{-15}{7} \)
\( A_5 = 15 + 5d = 15 + \frac{-150}{7} = \frac{105 - 150}{7} = \frac{-45}{7} \)
\( A_6 = 15 + 6d = 15 + \frac{-180}{7} = \frac{105 - 180}{7} = \frac{-75}{7} \)

Hence, required A.M.s between 15 and -15 = \( \frac{75}{7}, \frac{45}{7}, \frac{15}{7}, \frac{-15}{7}, \frac{-45}{7} \) and \( \frac{-75}{7} \)
In simple words: We made 8 equally spaced numbers from 15 to -15. The gap is 30, divided into 7 equal parts. Each step down is 30/7.

📝 Teacher's Note: Show students the pattern: the numerators go 75, 45, 15, -15, -45, -75. They decrease by 30 each time. This helps check the arithmetic.

🎯 Exam Tip: When inserting 6 means, you get 8 terms total. The last term is the 8th term, so use a + 7d. Verify by checking that A_6 + d = -15.

Exercise 10E

Question 1. Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km hr⁻¹ The second car goes at a speed of 8 km h⁻¹ in the first hour and thereafter increasing the speed by 0.5 km h⁻¹ each succeeding hour. After how many hours will the two cars meet?
Answer:
Given:
First car speed = 10 km/hr (constant)
Second car speed = 8 km/hr in first hour, then increases by 0.5 km/hr each hour

Solution:
Let the two cars meet after n hours.
That means the two cars travel the same distance in n hours.

Distance travelled by the 1ˢᵗ car in n hours = \( 10 \times n \) km

Distance travelled by the 2ⁿᵈ car in n hours = \( \frac{n}{2}[2 \times 8 + (n - 1) \times 0.5] \) km

\( \Rightarrow 10 \times n = \frac{n}{2}[2 \times 8 + (n - 1) \times 0.5] \)

\( \Rightarrow 20 = 16 + 0.5n - 0.5 \)
\( \Rightarrow 0.5n = 4.5 \)
\( \Rightarrow n = 9 \)

Thus, the two cars will meet after 9 hours.
In simple words: We set up an equation where both cars travel the same distance. The first car goes at steady speed. The second car speeds up each hour in an arithmetic pattern. We solve to find when they meet.

📝 Teacher's Note: Show students that the second car's speeds form an AP: 8, 8.5, 9, 9.5... Use the sum of AP formula for distance. Draw a simple diagram showing both cars starting together.

🎯 Exam Tip: Always write "Let the cars meet after n hours" first. Set distance equal for both cars. Use the AP sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \) for the second car's distance.

Question 2. A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes.
Answer:
Given:
Total amount of prize = \( S_n \) = Rs. 700
Number of prizes = n = 7
Depreciation in next prize = -Rs. 20

Solution:
Let the value of the first prize be Rs. a.
Let the value of first prize be Rs. a.
Depreciation in next prize = -Rs. 20
We have,
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)

\( \Rightarrow 700 = \frac{7}{2}[2a + 6(-20)] \)

\( \Rightarrow 700 = \frac{7}{2}[2a - 120] \)

\( \Rightarrow 1400 = 14a - 840 \)
\( \Rightarrow 14a = 2240 \)
\( \Rightarrow a = 160 \)

\( \Rightarrow \) Value of 1ˢᵗ prize = Rs. 160
Value of 2ⁿᵈ prize = Rs. (160 - 20) = Rs. 140
Value of 3ʳᵈ prize = Rs. (140 - 20) = Rs. 120
Value of 4ᵗʰ prize = Rs. (120 - 20) = Rs. 100
Value of 5ᵗʰ prize = Rs. (100 - 20) = Rs. 80
Value of 6ᵗʰ prize = Rs. (80 - 20) = Rs. 60
Value of 7ᵗʰ prize = Rs. (60 - 20) = Rs. 40

In simple words: The prizes decrease by Rs. 20 each time. We use the sum formula to find the first prize, then subtract Rs. 20 for each next prize.

📝 Teacher's Note: Emphasize that this is a decreasing AP with d = -20. Students often forget the negative sign. Show them how to check: add all seven prizes to get Rs. 700.

🎯 Exam Tip: Write d = -20 clearly at the start. List all seven prize values at the end. Always verify that the sum equals Rs. 700 to check your answer.

Question 3. An article can be bought by paying Rs. 28,000 at once or by making 12 monthly instalments. If the first instalment paid is Rs. 3,000 and every other instalment is Rs. 100 less than the previous one, find :
(i) amount of instalment paid in the 9th month
(ii) total amount paid in the instalment scheme.
Answer:
Given:
Number of instalments = n = 12
First instalment = a = Rs. 3000
Depreciation in instalment = d = -100

(i) Amount of instalment paid in the 9th month
= \( t_9 \)
= a + 8d
= 3000 + 8 × (-100)
= 3000 - 800
= Rs. 2200

(ii) Total amount paid in the instalment scheme
= \( S_{12} \)
= \( \frac{12}{2}[2 \times 3000 + 11 \times (-100)] \)
= 6[6000 - 1100]
= 6 × 4900
= Rs. 29,400

In simple words: Each month the payment goes down by Rs. 100. For the 9th month, we subtract Rs. 800 from the first payment. The total of all 12 payments is Rs. 29,400.

📝 Teacher's Note: Show students that paying in instalments costs Rs. 1,400 more than paying at once (29,400 - 28,000 = 1,400). This is like interest. Use the formula \( t_n = a + (n-1)d \) for individual terms.

🎯 Exam Tip: For part (i), use \( t_9 = a + 8d \) (not 9d). For part (ii), use the sum formula. Always state what each part is asking for before solving.

Question 4. A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year.
Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10th year.
(iii) the total production in 7 years.
Answer:
Given:
Since the production increases uniformly by a fixed number every year, the sequence formed by the production in different years is an A.P.
Let the production in the first year = a
Common difference = Number of units by which the production increases every year = d

We have,
\( t_3 = 600 \)
\( \Rightarrow a + 2d = 600 \) .....(i)
\( t_7 = 700 \)
\( \Rightarrow a + 6d = 700 \) .....(ii)

Subtracting (i) from (ii), we get
4d = 100 ⟹ d = 25
\( \Rightarrow a + 2 \times 25 = 600 \)
\( \Rightarrow a = 550 \)

(i) The production in the first year = 550 TV sets
(ii) Production in the 10th year = \( t_{10} = 550 + 9 \times 25 = 775 \) TV sets
(iii) Production in 7 years = \( S_7 = \frac{7}{2}[2 \times 550 + 6 \times 25] \)
= \( \frac{7}{2}[1100 + 150] \)
= \( \frac{7}{2} \times 1250 \)
= 4375 TV sets

In simple words: Production goes up by 25 units each year. We use two given years to find the starting production and the yearly increase. Then we can find any year's production.

📝 Teacher's Note: Show students how to set up two equations with the given information. Emphasize that subtracting equations eliminates 'a' to find 'd'. Draw a timeline showing years 1, 3, 7, and 10.

🎯 Exam Tip: Always write the two equations clearly first. Show the subtraction step to find d. For part (ii), remember it's \( t_{10} = a + 9d \), not 10d. Check: year 3 should give 600 units.

Question 5. Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying instalments every month. If the instalment for the first month is Rs. 1,000 and it increases by Rs. 100 every month, what amount will she pay as the 30th instalment of loan? What amount of loan she still has to pay after the 30th instalment?
Answer:
Given:
Total amount of loan = Rs. 1,18,000
First instalment = a = Rs. 1000
Increase in instalment every month = d = Rs. 100

30th instalment = \( t_{30} \)
= a + 29d
= 1000 + 29 × 100
= 1000 + 2900
= Rs. 3900

Now, amount paid in 30 instalments = \( S_{30} \)
= \( \frac{30}{2}[2 \times 1000 + 29 \times 100] \)
= 15[2000 + 2900]
= 15 × 4900
= Rs. 73,500

∴ Amount of loan to be paid after the 30th instalments
= Rs. (1,18,000 - 73,500)
= Rs. 44,500

In simple words: Each month Mrs. Gupta pays Rs. 100 more than the previous month. The 30th payment is Rs. 3,900. After 30 payments, she still owes Rs. 44,500.

📝 Teacher's Note: Help students understand that the instalments increase each month (unlike Question 3 where they decreased). Show how to subtract total paid from original loan to find remaining amount.

🎯 Exam Tip: For the 30th term, use \( t_{30} = a + 29d \). Calculate sum of 30 terms, then subtract from total loan. Always show the final subtraction step clearly.

Question 6. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?
Answer:
Given:
Since each section of each class plants five times the number of trees as the class number and there are three sections of each class, we have:
Total number of trees planted by the students from class 1 to 10
= 3[1 × 5 + 2 × 5 + 3 × 5 + ...... + 10 × 5]
= 3[5 + 10 + 15 + ...... + 50]
= 3[\( \frac{10}{2}(2 \times 5 + 9 \times 5) \)]
= 3[5(10 + 45)]
= 3 × 5 × 55
= 825

Hence, 825 trees were planted by students.

In simple words: Class 1 plants 5 trees per section, Class 2 plants 10 trees per section, and so on. Each class has 3 sections. So we multiply by 3 at the end.

📝 Teacher's Note: Explain that the sequence is 5, 10, 15, 20, ..., 50 (which is 5 times 1, 2, 3, ..., 10). Each term represents trees per section. Since there are 3 sections per class, multiply the total by 3.

🎯 Exam Tip: First write the sequence: 5, 10, 15, ..., 50. Use AP sum formula with a = 5, d = 5, n = 10. Don't forget to multiply by 3 for the three sections per class.

Exercise 10F

Question 1. The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Answer:
Given:
6th term = 16
14th term = 32

Step 1: Let 'a' be the first term and 'd' be the common difference of the given A.P.
t₆ = 16
\( \implies \) a + 5d = 16 ….(i)

Step 2: And t₁₄ = 32
\( \implies \) a + 13d = 32 ….(ii)

Step 3: Subtracting (i) from (ii), we get
8d = 16
\( \implies \) d = 2

Step 4: From equation (i):
a + 5(2) = 16
\( \implies \) a = 6

Step 5: Hence, 36th term = t₃₆ = a + 35d = 6 + 35(2) = 76
In simple words: We found the first term and common difference first. Then we used the formula to find the 36th term. It is 76.

📝 Teacher's Note: Show students to always set up two equations first. One equation is not enough to find two unknown values (a and d). Solving two equations together gives the answer.

🎯 Exam Tip: Always write "Given" first, then set up equations clearly. Show all steps. Final answer must be clearly stated at the end.

Question 2. If the third and the 9th terms of an A.P. are 4 and -8, find which term of this A.P. is 0.
Answer:
Given:
t₃ = 4
t₉ = -8

Step 1: For an A.P.,
t₃ = 4
\( \implies \) a + 2d = 4 … (i)
t₉ = -8
\( \implies \) a + 8d = -8 …. (ii)

Step 2: Subtracting (i) from (ii), we get
6d = -12
\( \implies \) d = -2

Step 3: Substituting d = -2 in (i), we get
a + 2(-2) = 4
\( \implies \) a - 4 = 4
\( \implies \) a = 8

Step 4: General term = tₙ = 8 + (n - 1)(-2)
Let pth term of this A.P. be 0.
8 + (p - 1)(-2) = 0
8 - 2p + 2 = 0
10 - 2p = 0
2p = 10
\( \implies \) p = 5

Thus, 5th term of this A.P. is 0.
In simple words: We found the pattern of the A.P. first. Then we put the value 0 in the formula to find which position gives us 0.

📝 Teacher's Note: Explain that when we want to find which term equals a given value, we substitute that value into the general term formula. Then solve for n.

🎯 Exam Tip: Write the general term formula clearly. Then substitute the given value (here 0) and solve step by step. Always state your final answer clearly.

Question 3. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Answer:
Given:
Number of terms, n = 50
3rd term, t₃ = 12
Last term, l = 106

Step 1: From the given information:
a + 2d = 12 ….(i)
t₅₀ = 106
\( \implies \) a + 49d = 106 ….(ii)

Step 2: Subtracting (i) from (ii), we get
47d = 94
\( \implies \) d = 2

Step 3: From equation (i):
a + 2(2) = 12
\( \implies \) a = 8

Step 4: Hence, t₂₉ = a + 28d = 8 + 28(2) = 8 + 56 = 64
In simple words: We used the 3rd term and last term to find the first term and common difference. Then we calculated the 29th term using the formula.

📝 Teacher's Note: Show students that the last term of an A.P. with n terms is tₙ = a + (n-1)d. Here, t₅₀ = a + 49d because there are 50 terms.

🎯 Exam Tip: For last term, remember the formula is a + (n-1)d, not a + nd. Be careful with the number of terms and term position.

Question 4. Find the arithmetic mean of:
(i) -5 and 41
(ii) 3x - 2y and 3x + 2y
(iii) (m + n)² and (m - n)²
Answer:
(i) Arithmetic mean of -5 and 41 = \( \frac{-5 + 41}{2} = \frac{36}{2} = 18 \)

(ii) Arithmetic mean of (3x - 2y) and (3x + 2y) = \( \frac{3x - 2y + 3x + 2y}{2} = \frac{6x}{2} = 3x \)

(iii) Arithmetic mean of (m + n)² and (m - n)² = \( \frac{(m + n)^2 + (m - n)^2}{2} \)
= \( \frac{m^2 + n^2 + 2mn + m^2 + n^2 - 2mn}{2} \)
= \( \frac{2(m^2 + n^2)}{2} \)
= m² + n²

In simple words: Arithmetic mean is just the average of two numbers. Add them and divide by 2. For algebraic expressions, we follow the same rule.

📝 Teacher's Note: Arithmetic mean formula is very simple: (a + b)/2. For algebraic expressions, expand carefully and simplify step by step. Show students how terms cancel out in part (ii).

🎯 Exam Tip: Always write the formula first: Arithmetic mean = (sum of terms)/(number of terms). For two numbers, it is (a + b)/2. Show all algebraic steps clearly.

Question 5. Find the sum of first 10 terms of the A.P. 4 + 6 + 8 + ………
Answer:
Given:
First term, a = 4
Common difference, d = 6 - 4 = 2
n = 10

Step 1: Using the formula \( S = \frac{n}{2}[2a + (n - 1)d] \)
= \( \frac{10}{2}[2(4) + (10 - 1)(2)] \)
= 5[8 + 18]
= 5 × 26
= 130

In simple words: We used the sum formula for A.P. The sum of first 10 terms is 130.

📝 Teacher's Note: There are two formulas for sum: Sₙ = n/2[2a + (n-1)d] or Sₙ = n/2(first term + last term). Use the first formula when you know a and d.

🎯 Exam Tip: Always identify a, d, and n first. Write the sum formula clearly. Substitute values step by step. Don't skip steps in calculation.

Question 6. Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.
Answer:
Given:
First term, a = 3
Last term, l = 57
n = 20

Step 1: Using the formula \( S = \frac{n}{2}(a + l) \)
= \( \frac{20}{2}(3 + 57) \)
= 10 × 60
= 600

In simple words: When we know the first and last terms, we use a simpler formula. Just add them and multiply by half the number of terms.

📝 Teacher's Note: When you have first term and last term, use S = n/2(a + l). This is much easier than the other formula. Students should learn both formulas.

🎯 Exam Tip: Choose the easier formula. If you have first and last terms, use S = n/2(a + l). It saves time and reduces calculation errors.

Question 7. How many terms of the series 18 + 15 + 12 + ……. when added together will give 45?
Answer:
Step 1: Check if it's an A.P.
15 - 18 = 12 - 15 = -3
Thus, the given series is an A.P. with first term 18 and common difference -3.

Step 2: Let the number of terms to be added be 'n'.
Using \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
45 = \( \frac{n}{2}[2(18) + (n - 1)(-3)] \)
90 = n[36 - 3n + 3]
90 = n[39 - 3n]
90 = 3n[13 - n]
30 = 13n - n²
\( \implies \) n² - 13n + 30 = 0

Step 3: Factoring the quadratic:
n² - 10n - 3n + 30 = 0
n(n - 10) - 3(n - 10) = 0
(n - 10)(n - 3) = 0
\( \implies \) n = 10 or n = 3

Thus, required number of terms to be added is 3 or 10.
In simple words: We got two answers because both 3 terms and 10 terms can give the same sum. This happens in A.P. sometimes when we have negative common difference.

📝 Teacher's Note: When common difference is negative, the sum increases first then decreases. That's why we can get the same sum with different number of terms.

🎯 Exam Tip: Always check your quadratic equation carefully. Both solutions can be valid in A.P. problems. Write both answers unless the question asks for one specific case.

Question 8. The nth term of a sequence is 8 - 5n. Show that the sequence is an A.P.
Answer:
Given: tₙ = 8 - 5n

Step 1: Find the (n+1)th term:
Replacing n by (n + 1), we get
tₙ₊₁ = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n

Step 2: Find the common difference:
tₙ₊₁ - tₙ = (3 - 5n) - (8 - 5n) = -5

Step 3: Since (tₙ₊₁ - tₙ) is independent of n and is therefore a constant, the given sequence is an A.P.

In simple words: To prove a sequence is A.P., we show that the difference between consecutive terms is always the same. Here it's always -5.

📝 Teacher's Note: To prove any sequence is A.P., find tₙ₊₁ - tₙ. If this difference is constant (same number always), then it's an A.P.

🎯 Exam Tip: Write "tₙ₊₁ - tₙ = constant" to show it's an A.P. Always state clearly that the difference is independent of n.

Question 9. Find the general term (nth term) and 23rd term of the sequence 3, 1, -1, -3, ……
Answer:
Step 1: Check if it's an A.P.
1 - 3 = -1 - 1 = -3 - (-1) = -2
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.

Step 2: Find the general term:
The general term (nth term) of an A.P. is given by
tₙ = a + (n - 1)d
= 3 + (n - 1)(-2)
= 3 - 2n + 2
= 5 - 2n

Step 3: Find the 23rd term:
t₂₃ = 5 - 2(23) = 5 - 46 = -41

In simple words: The formula for any term in this sequence is 5 - 2n. The 23rd term is -41.

📝 Teacher's Note: Always check that it's an A.P. first by finding the common difference. Then use the formula tₙ = a + (n-1)d to get the general term.

🎯 Exam Tip: Write the general term formula clearly. Then substitute the specific value of n to find any particular term. Show all substitution steps.

Question 10. Which term of the sequence 3, 8, 13, …….. is 78?
Answer:
Step 1: Check if it's an A.P.
8 - 3 = 13 - 8 = 5
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.

Step 2: Let the nth term of the given A.P. be 78.
tₙ = 78
a + (n - 1)d = 78
3 + (n - 1)(5) = 78
3 + 5n - 5 = 78
5n - 2 = 78
5n = 80
n = 16

Thus, the 16th term of the given sequence is 78.
In simple words: We put 78 in the general term formula and solved for n. The answer is the 16th term.

📝 Teacher's Note: When asked "which term equals a given value", substitute that value into the general term formula and solve for n.

🎯 Exam Tip: Set up the equation tₙ = given value, then solve for n step by step. Always check your answer by substituting back.

Question 11. Is -150 a term of 11, 8, 5, 2, ……… ?
Answer:
Step 1: Check if it's an A.P.
8 - 11 = 5 - 8 = 2 - 5 = -3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.

Step 2: Let the nth term be -150.
tₙ = -150
a + (n - 1)d = -150
11 + (n - 1)(-3) = -150
11 - 3n + 3 = -150
14 - 3n = -150
-3n = -164
n = \( \frac{164}{3} = 54.67 \)

Step 3: Since n is not a whole number, -150 is not a term of this sequence.

In simple words: For a number to be a term of an A.P., the position (n) must be a whole number. Since we got a decimal, -150 is not in this sequence.

📝 Teacher's Note: For any number to be a term of a sequence, n must be a positive integer. If you get a fraction or decimal, that number is not in the sequence.

🎯 Exam Tip: Always check if your value of n is a positive integer. If not, clearly state that the given number is not a term of the sequence.

Question 12. How many two digit numbers are divisible by 3?
Answer: The two-digit numbers divisible by 3 are as follows: 12, 15, 18, 21, …….. 99
Clearly, this forms an A.P. with first term, a = 12
and common difference, d = 3
Last term = nth term = 99
The general term of an A.P. is given by
\( t_n = a + (n - 1)d \)
\( \implies 99 = 12 + (n - 1)(3) \)
\( \implies 99 = 12 + 3n - 3 \)
\( \implies 90 = 3n \)
\( \implies n = 30 \)
Thus, 30 two-digit numbers are divisible by 3.
In simple words: We find all two-digit numbers that can be divided by 3. They make a pattern: 12, 15, 18... We count how many terms are in this pattern.

📝 Teacher's Note: Show students the pattern: start with 12, keep adding 3. Common mistake is starting with 10 or 11 instead of 12 (first two-digit number divisible by 3).

🎯 Exam Tip: Always identify first term correctly. Write a = 12, d = 3, last term = 99. Show all steps clearly to get full marks.

Question 13. How many multiples of 4 lie between 10 and 250?
Answer: Numbers between 10 and 250 which are multiple of 4 are as follows: 12, 16, 20, 24,……, 248
Clearly, this forms an A.P. with first term a = 12,
common difference d = 4 and last term l = 248
\( l = a + (n - 1)d \)
\( \implies 248 = 12 + (n - 1) \times 4 \)
\( \implies 236 = (n - 1) \times 4 \)
\( \implies n - 1 = 59 \)
\( \implies n = 60 \)
Thus, 60 multiples of 4 lie between 10 and 250.
In simple words: We list all numbers between 10 and 250 that can be divided by 4. They form a pattern starting with 12. We count 60 such numbers.

📝 Teacher's Note: First multiple of 4 after 10 is 12 (not 10). Last multiple of 4 before 250 is 248. Students often include 10 or 250 by mistake.

🎯 Exam Tip: Write "between" means not including 10 and 250. Find correct first and last terms. Show step-by-step calculation.

Question 14. The sum of the 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th term is 44. Find the first three terms of the A.P.
Answer:
Given, \( t_4 + t_8 = 24 \)
\( (a + 3d) + (a + 7d) = 24 \)
\( 2a + 10d = 24 \)
\( a + 5d = 12 \) ....(i)
And,
\( t_6 + t_{10} = 44 \)
\( (a + 5d) + (a + 9d) = 44 \)
\( 2a + 14d = 44 \)
\( a + 7d = 22 \) ...(ii)
Subtracting (i) from (ii), we get
\( 2d = 10 \)
\( d = 5 \)
Substituting value of d in (i), we get
\( a + 5 \times 5 = 12 \)
\( a + 25 = 12 \)
\( a = -13 \) = 1st term
\( a + d = -13 + 5 = -8 \) = 2nd term
\( a + 2d = -13 + 2 \times 5 = -13 + 10 = -3 \) = 3rd term
Hence, the first three terms of an A.P. are -13, -8 and -3.
In simple words: We use two given conditions to make two equations. We solve these equations to find a and d. Then we calculate the first three terms.

📝 Teacher's Note: Make two equations from given information. Solve by elimination method. Common mistake is wrong calculation when subtracting equations.

🎯 Exam Tip: Write "Given" clearly. Make two equations and label them (i) and (ii). Show subtraction step. Calculate all three terms at the end.

Question 15. The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Answer:
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Given,
\( S_{14} = 1050 \)
\( \frac{14}{2}[2a + 13d] = 1050 \)
\( 7[2a + 13d] = 1050 \)
\( 2a + 13d = 150 \)
\( a + 6.5d = 75 \) ....(i)
And, \( t_{14} = 140 \)
\( a + 13d = 140 \) ....(ii)
Subtracting (i) from (ii), we get
\( 6.5d = 65 \)
\( d = 10 \)
\( a + 13(10) = 140 \)
\( a = 10 \)
Thus, 20th term = \( t_{20} = 10 + 19d = 10 + 19(10) = 200 \)
In simple words: We use sum formula and nth term formula to make two equations. We solve to find a and d. Then we find the 20th term.

📝 Teacher's Note: Use sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \) and nth term formula. Students often confuse which formula to use when.

🎯 Exam Tip: Write both formulas clearly. Make two equations from given data. Show all calculation steps. Write final answer clearly.

Question 16. The 25th term of an A.P. exceeds its 9th term by 16. Find its common difference.
Answer:
nth term of an A.P. is given by \( t_n = a + (n - 1)d \).
\( t_{25} = a + (25 - 1)d = a + 24d \) and
\( t_9 = a + (9 - 1)d = a + 8d \)
According to the condition in the question, we get
\( t_{25} = t_9 + 16 \)
\( a + 24d = a + 8d + 16 \)
\( 16d = 16 \)
\( d = 1 \)
In simple words: The 25th term is 16 more than the 9th term. We write this as an equation and solve to find the common difference.

📝 Teacher's Note: "Exceeds by" means "is more than by that amount." Write the equation carefully. The 'a' terms cancel out nicely.

🎯 Exam Tip: Write both terms using nth term formula. Set up equation correctly: larger term = smaller term + difference. Solve step by step.

Question 17. For an A.P., show that: (m + n)th term + (m - n)th term = 2 × mth term
Answer:
Let a and d be the first term and common difference respectively.
\( (m + n)^{th} \) term = \( a + (m + n - 1)d \) .... (i) and
\( (m - n)^{th} \) term = \( a + (m - n - 1)d \) .... (ii)
From (i) + (ii), we get
(m + n)th term + (m - n)th term
= \( a + (m + n - 1)d + a + (m - n - 1)d \)
= \( a + md + nd - d + a + md - nd - d \)
= \( 2a + 2md - 2d \)
= \( 2a + (m - 1)2d \)
= \( 2[a + (m - 1)d] \)
= 2 × mth term
Hence proved.
In simple words: We write the two terms using the nth term formula. We add them and simplify. The result equals 2 times the mth term.

📝 Teacher's Note: This is a proof question. Write each term clearly using formula. Add carefully and group like terms. Show every algebraic step.

🎯 Exam Tip: Write "Let a and d be..." at start. Show addition step clearly. Group nd and -nd terms (they cancel). End with "Hence proved."

Question 18. If the nth term of the A.P. 58, 60, 62,.... is equal to the nth term of the A.P. -2, 5, 12, ...., find the value of n.
Answer:
In the first A.P. 58, 60, 62,....
a = 58 and d = 2
\( t_n = a + (n - 1)d \)
\( t_n = 58 + (n - 1)2 \) .... (i)
In the second A.P. -2, 5, 12, ....
a = -2 and d = 7
\( t_n = a + (n - 1)d \)
\( t_n = -2 + (n - 1)7 \) .... (ii)
Given that the nth term of first A.P is equal to the nth term of the second A.P.
\( 58 + (n - 1)2 = -2 + (n - 1)7 \) ... from (i) and (ii)
\( 58 + 2n - 2 = -2 + 7n - 7 \)
\( 56 + 2n = -9 + 7n \)
\( 65 = 5n \)
\( n = 13 \)
In simple words: We find the nth term formula for both A.P.s. We set them equal and solve to find which term number (n) makes them equal.

📝 Teacher's Note: Find first term and common difference for both A.P.s separately. Write nth term for each. Set equal and solve the equation.

🎯 Exam Tip: Write a and d for both A.P.s clearly. Show nth term formulas. Set them equal and expand brackets carefully before solving.

Question 19. Which term of the A.P. 105, 101, 97 … is the first negative term?
Answer:
Here a = 105 and d = 101 - 105 = -4
Let \( a_n \) be the first negative term.
\( a_n < 0 \)
\( a + (n - 1)d < 0 \)
\( 105 + (n - 1)(-4) < 0 \)
\( 105 - 4n + 4 < 0 \)
\( 109 - 4n < 0 \)
\( 109 < 4n \)
\( n > 27.25 \)
Since n must be a whole number, the smallest value is n = 28.
Therefore, the 28th term is the first negative term.
In simple words: The numbers keep getting smaller by 4 each time. We find which term number makes the value negative for the first time.

📝 Teacher's Note: Common difference is negative here. Set up inequality \( t_n < 0 \). Solve to find when term becomes negative. Take next whole number.

🎯 Exam Tip: Find d correctly (it's negative). Set nth term less than 0. Solve inequality. Since n must be whole number, round up to next integer.

Question 20. How many three digit numbers are divisible by 7?
Answer:
The first three digit number which is divisible by 7 is 105 and the last three digit number which is divisible by 7 is 994.
This is an A.P. in which a = 105, d = 7 and \( t_n = 994 \).
We know that nth term of A.P is given by
\( t_n = a + (n - 1)d \)
\( 994 = 105 + (n - 1)7 \)
\( 889 = 7n - 7 \)
\( 896 = 7n \)
\( n = 128 \)
∴ There are 128 three digit numbers which are divisible by 7.
In simple words: We find the first and last three-digit numbers that can be divided by 7. They form a pattern. We count how many numbers are in this pattern.

📝 Teacher's Note: First three-digit multiple of 7 is 105 (not 100). Last is 994 (not 999). Students often pick wrong endpoints.

🎯 Exam Tip: Check: 105 ÷ 7 = 15, 994 ÷ 7 = 142. Write a = 105, d = 7, last term = 994. Show all steps clearly.

Question 21. Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.
Answer:
Let the three parts of 216 in A.P be (a - d), a, (a + d).
\( a - d + a + a + d = 216 \)
\( 3a = 216 \)
\( a = 72 \)
Given that the product of the two smaller parts is 5040.
\( a(a - d) = 5040 \)
\( 72(72 - d) = 5040 \)
\( 72 - d = 70 \)
\( d = 2 \)
∴ a - d = 72 - 2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.
In simple words: We write three numbers in A.P. as (a-d), a, (a+d). Their sum is 216, so we find a. Then we use the product condition to find d.

📝 Teacher's Note: Use (a-d), a, (a+d) format for three terms in A.P. The middle term is always 'a'. Product of two smaller means (a-d) and a.

🎯 Exam Tip: Write three terms as (a-d), a, (a+d). Use sum = 216 to find a first. Then use product condition to find d. Calculate all three parts.

Question 22. Can 2n² - 7 be the nth term of an A.P? Explain.
Answer:
We have 2n² - 7,
Substitute n = 1, 2, 3, …, we get
2(1)² - 7, 2(2)² - 7, 2(3)² - 7, 2(4)² - 7, ....
-5, 1, 11, 25, ....
Difference between the first and second term = 1 - (-5) = 6
And Difference between the second and third term = 11 - 1 = 10
Here, the common difference is not same.
Therefore the nth term of an A.P can't be 2n² - 7.
In simple words: We find the first few terms using the given formula. We check if the difference between consecutive terms is the same. It's not, so this can't be an A.P.

📝 Teacher's Note: In A.P., difference between consecutive terms must be constant. Here differences are 6, 10, 14... (not constant). So it's not A.P.

🎯 Exam Tip: Calculate first few terms. Find differences between consecutive terms. If differences are not equal, it's not an A.P. Show calculation clearly.

Question 23. Find the sum of the A.P., 14, 21, 28, …, 168.
Answer:
Here a = 14, d = 7 and \( t_n = 168 \)
\( t_n = a + (n - 1)d \)
\( 168 = 14 + (n - 1)7 \)
\( 154 = 7n - 7 \)
\( 161 = 7n \)
\( n = 23 \)
We know that,
\( S_n = \frac{n}{2}(a + l) \)
\( S_{23} = \frac{23}{2}(14 + 168) \)
\( S_{23} = \frac{23}{2} \times 182 \)
\( S_{23} = 23 \times 91 = 2093 \)
In simple words: First we find how many terms are there (n = 23). Then we use the sum formula with first term, last term, and number of terms.

📝 Teacher's Note: First find n using nth term formula. Then use sum formula \( S_n = \frac{n}{2}(a + l) \) where l is last term. Don't forget to find n first.

🎯 Exam Tip: Find n first using \( t_n = a + (n-1)d \). Then use sum formula \( S_n = \frac{n}{2}(a + l) \). Show both steps clearly for full marks.

Question 24. The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31st term of this A.P.
Answer:
Given:
First term (a) = 20
Sum of first seven terms (\( S_7 \)) = 2100

Step 1: Find the common difference (d) using the sum formula.
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)

Step 2: Substitute the given values.
\( S_7 = \frac{7}{2}[2 \times 20 + (7 - 1)d] \)
\( 2100 = \frac{7}{2}(40 + 6d) \)

Step 3: Solve for d.
\( 2100 = \frac{7}{2}(40 + 6d) \)
\( 4200 = 7(40 + 6d) \)
\( 600 = 40 + 6d \)
\( d = \frac{560}{6} \)

Step 4: Find the 31st term using the formula.
\( t_n = a + (n - 1)d \)
\( t_{31} = 20 + (31 - 1) \times \frac{560}{6} \)
\( t_{31} = 20 + 30 \times \frac{560}{6} \)
\( t_{31} = 20 + 5 \times 560 \)
\( t_{31} = 20 + 2800 = 2820 \)

Therefore the 31st term of the given A.P. is 2820.
In simple words: We first found how much the sequence increases each time (common difference). Then we used that to find the 31st number in the sequence. It's like climbing stairs - we know the step size and count 31 steps up.

📝 Teacher's Note: Show students that we need to find 'd' first before finding any term. Draw a number line to show how A.P. works like equal steps. Common mistake is forgetting to simplify fractions.

🎯 Exam Tip: Always write "Given:" first and list all values. Show each step clearly. Write the final answer in a box or underline it. You get marks for method even if calculation is wrong.

Question 25. Find the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58.
Answer:
Step 1: Reverse the A.P. to find the sum of last 8 terms.
Original A.P.: -12, -10, -8, ……, 58
Reversed A.P.: 58, ……, -8, -10, -12

Step 2: Identify the first term and common difference of reversed A.P.
Here a = 58, d = -2

Step 3: Use the sum formula for first 8 terms of reversed A.P.
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)

Step 4: Substitute the values.
\( S_8 = \frac{8}{2}[2 \times 58 + (8 - 1)(-2)] \)
\( S_8 = 4[116 + 7(-2)] \)
\( S_8 = 4[116 - 14] \)
\( S_8 = 4 \times 102 = 408 \)

Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.
In simple words: To find the sum of last 8 terms, we flip the sequence backwards. Then we find the sum of first 8 terms of the flipped sequence. It's like reading a book from the end.

📝 Teacher's Note: Explain that "last 8 terms" means we start from the end. Show students how reversing makes the calculation easier. Use colored chalk to mark the last 8 terms clearly.

🎯 Exam Tip: When finding sum of last terms, always reverse the A.P. first. Write "Reversed A.P.:" clearly. Show the new first term and common difference. This method gets full marks.