OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Exercise 9 (B)

Question 1. Find the sum of the:
(i) first 15 terms of the A.P: 2, 5, 8, 11, ...
(ii) first 50 terms of the A.P: -27, -23, -19, ...
(iii) first 17 terms of the A.P: \( \frac { 1 }{ 5 } \), \( \frac { -3 }{ 10 } \), \( \frac { -4 }{ 5 } \) ...
(iv) first 24 terms of A.P: 0.6, 1.7, 2.8, ...
Answer:
(i) For the A.P: 2, 5, 8, 11, ...
Here, the first term \( a = 2 \). The common difference \( d = 5 - 2 = 3 \). We need the sum of the first 15 terms, so \( n = 15 \).
The formula for the sum of n terms of an A.P. is \( S_n = \frac{n}{2}[2a+(n-1)d] \).
Substitute the values:
\( S_{15} = \frac{15}{2}[2 \times 2+(15-1) \times 3] \)
\( = \frac{15}{2}[4+14 \times 3] \)
\( = \frac{15}{2}[4+42] \)
\( = \frac{15}{2}[46] \)
\( = 15 \times 23 \)
\( = 345 \)

(ii) For the A.P: -27, -23, -19, ...
Here, the first term \( a = -27 \). The common difference \( d = -23 - (-27) = -23 + 27 = 4 \). We need the sum of the first 50 terms, so \( n = 50 \).
Using the formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_{50} = \frac{50}{2}[2 \times(-27)+(50-1) \times 4] \)
\( = 25[-54 + 49 \times 4] \)
\( = 25[-54 + 196] \)
\( = 25[142] \)
\( = 3550 \)

(iii) For the A.P: \( \frac { 1 }{ 5 } \), \( \frac { -3 }{ 10 } \), \( \frac { -4 }{ 5 } \) ...
Here, the first term \( a = \frac { 1 }{ 5 } \). The common difference \( d = \frac { -3 }{ 10 } - \frac { 1 }{ 5 } = \frac{-3-2}{10} = \frac{-5}{10} = -\frac{1}{2} \). We need the sum of the first 17 terms, so \( n = 17 \).
Using the formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_{17} = \frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(-\frac{1}{2}\right)\right] \)
\( = \frac{17}{2}\left[\frac{2}{5}+16\left(-\frac{1}{2}\right)\right] \)
\( = \frac{17}{2}\left[\frac{2}{5}-8\right] \)
\( = \frac{17}{2}\left[\frac{2-40}{5}\right] \)
\( = \frac{17}{2}\left[\frac{-38}{5}\right] \)
\( = \frac{-323}{5} \)
\( = -64\frac{3}{5} \)

(iv) For the A.P: 0.6, 1.7, 2.8, ...
Here, the first term \( a = 0.6 \). The common difference \( d = 1.7 - 0.6 = 1.1 \). We need the sum of the first 24 terms, so \( n = 24 \).
Using the formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_{24} = \frac{24}{2}[2 \times 0.6 + (24-1)(1.1)] \)
\( = 12[1.2 + 23 \times 1.1] \)
\( = 12[1.2 + 25.3] \)
\( = 12 \times 26.5 \)
\( = 318.0 \)
In simple words: To find the sum of terms in an Arithmetic Progression, first identify the starting term, the common difference between terms, and how many terms you need to sum. Then, use the special sum formula \( S_n = \frac{n}{2}(2a + (n-1)d) \) by putting in your numbers. Each part of the problem needs these steps.

🎯 Exam Tip: Always identify the first term (a), common difference (d), and number of terms (n) correctly from the given A.P. before applying the sum formula. Pay close attention to negative signs and fractions to avoid calculation errors.

Question 2. Find the sum:
(i) 34 + 32 + 30 + ... + 2
(ii) \( 7 + 9\frac { 1 }{ 2 } + 12 + ... + 67 \).
Answer:
(i) For the series 34 + 32 + 30 + ... + 2:
This is an Arithmetic Progression (A.P.) where the first term \( a = 34 \).
The common difference \( d = 32 - 34 = -2 \).
The last term \( l = 2 \).
First, find the number of terms \( n \) using the formula for the nth term: \( l = a + (n-1)d \).
\( 2 = 34 + (n-1)(-2) \)
\( 2 - 34 = -2(n-1) \)
\( -32 = -2(n-1) \)
\( \implies \frac{-32}{-2} = n-1 \)
\( 16 = n-1 \)
\( n = 16 + 1 \)
\( n = 17 \)
Now, find the sum using the formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{17} = \frac{17}{2}(34+2) \)
\( = \frac{17}{2}(36) \)
\( = 17 \times 18 \)
\( = 306 \)

(ii) For the series \( 7 + 9\frac { 1 }{ 2 } + 12 + ... + 67 \):
This is an A.P. where the first term \( a = 7 \).
The common difference \( d = 9\frac { 1 }{ 2 } - 7 = \frac{19}{2} - 7 = \frac{19-14}{2} = \frac{5}{2} \).
The last term \( l = 67 \).
First, find the number of terms \( n \) using \( l = a + (n-1)d \).
\( 67 = 7 + (n-1)\frac{5}{2} \)
\( 67 - 7 = (n-1)\frac{5}{2} \)
\( 60 = (n-1)\frac{5}{2} \)
\( \implies (n-1) = \frac{60 \times 2}{5} \)
\( (n-1) = 12 \times 2 \)
\( n-1 = 24 \)
\( n = 24 + 1 \)
\( n = 25 \)
Now, find the sum using the formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{25} = \frac{25}{2}(7+67) \)
\( = \frac{25}{2}(74) \)
\( = 25 \times 37 \)
\( = 925 \)
In simple words: When you need to find the sum of numbers in an Arithmetic Progression (a list where each number increases or decreases by the same amount), first find out how many numbers there are. You can do this by using the first and last numbers, and the difference between them. Once you know how many numbers, use the sum formula that includes the first and last terms for an easy calculation.

🎯 Exam Tip: Remember to correctly identify 'a', 'd', and 'l' from the sequence. If 'n' is not given, always calculate it first before using the sum formula. The formula \( S_n = \frac{n}{2}(a+l) \) is particularly useful when the last term is known.

Question 3. The common difference of an A.P. is -2. Find its sum, if its first term is 100 and the last term is -10.
Answer:
In this Arithmetic Progression (A.P.):
The common difference \( d = -2 \).
The first term \( a = 100 \).
The last term \( l = -10 \).
First, we need to find the number of terms \( n \) in this A.P. We use the formula for the nth term: \( l = a + (n-1)d \).
Substitute the given values:
\( -10 = 100 + (n-1)(-2) \)
\( -10 - 100 = (n-1)(-2) \)
\( -110 = -2(n-1) \)
\( \implies \frac{-110}{-2} = n-1 \)
\( 55 = n-1 \)
\( n = 55 + 1 \)
\( n = 56 \)
Now that we have \( n \), we can find the sum of the A.P. using the formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{56} = \frac{56}{2}[100 + (-10)] \)
\( = 28[100 - 10] \)
\( = 28[90] \)
\( = 2520 \)
In simple words: To find the total sum of this number pattern, we first figure out how many numbers are in the list. We use the starting number, the ending number, and how much each number changes by. Once we know the count, we can easily add them all up using a quick sum rule.

🎯 Exam Tip: Always determine the number of terms 'n' first if it's not directly given. This can be found using the formula for the nth term \( a_n = a + (n-1)d \). Then use \( S_n = \frac{n}{2}(a+l) \) for efficient calculation.

Question 4. How many terms of the A.P. 54, 51, 48, ... should be taken so that their sum is 513?
Answer:
For the given A.P: 54, 51, 48, ...
The first term \( a = 54 \).
The common difference \( d = 51 - 54 = -3 \).
The sum of n terms \( S_n = 513 \).
We use the formula for the sum of n terms of an A.P.: \( S_n = \frac{n}{2}[2a+(n-1)d] \).
Substitute the known values:
\( 513 = \frac{n}{2}[2 \times 54 + (n-1)(-3)] \)
\( 513 = \frac{n}{2}[108 - 3n + 3] \)
Multiply both sides by 2 to remove the fraction:
\( 1026 = n[111 - 3n] \)
\( 1026 = 111n - 3n^2 \)
Rearrange the terms into a quadratic equation:
\( 3n^2 - 111n + 1026 = 0 \)
Divide the entire equation by 3 to simplify:
\( n^2 - 37n + 342 = 0 \)
Now, factorize the quadratic equation to find the values of \( n \). We need two numbers that multiply to 342 and add up to -37. These numbers are -18 and -19.
\( n^2 - 18n - 19n + 342 = 0 \)
Factor by grouping:
\( n(n - 18) - 19(n - 18) = 0 \)
\( (n - 18)(n - 19) = 0 \)
This gives two possible values for \( n \):
\( n - 18 = 0 \implies n = 18 \)
or
\( n - 19 = 0 \implies n = 19 \)
Both 18 and 19 are valid numbers of terms. This can happen in a decreasing A.P. where a later term might be zero, making the sum the same for n and n+1 terms.
In simple words: We are looking for how many numbers from the given pattern add up to 513. We use a special formula for sums and solve it like a puzzle to find two possible counts of numbers that both work. This shows that the sum can be the same for two different numbers of terms in a decreasing sequence if some terms are positive and some are negative, or zero.

🎯 Exam Tip: When solving for 'n' in a sum problem, you might get a quadratic equation. Remember to solve it and check both possible values of 'n'. Sometimes both answers are valid, and sometimes one might lead to an illogical term (like a negative number of logs in Question 19), which you should identify and discard.

Question 5. If the sum of first 9 terms of an A.P. is 72 and the common difference is 5, find the first term and the 10th term of the A.P.
Answer:
For the given Arithmetic Progression (A.P.):
The sum of the first 9 terms \( S_9 = 72 \).
The number of terms \( n = 9 \).
The common difference \( d = 5 \).
Let \( a \) be the first term.
We use the formula for the sum of n terms of an A.P.: \( S_n = \frac{n}{2}[2a+(n-1)d] \).
Substitute the known values:
\( 72 = \frac{9}{2}[2a + (9-1) \times 5] \)
\( 72 = \frac{9}{2}[2a + 8 \times 5] \)
\( 72 = \frac{9}{2}[2a + 40] \)
Multiply both sides by 2 and divide by 9:
\( \frac{72 \times 2}{9} = 2a + 40 \)
\( 8 \times 2 = 2a + 40 \)
\( 16 = 2a + 40 \)
\( 16 - 40 = 2a \)
\( -24 = 2a \)
\( \implies a = \frac{-24}{2} \)
\( a = -12 \)
So, the first term is -12.
Next, we need to find the 10th term, \( a_{10} \). We use the formula for the nth term: \( a_n = a + (n-1)d \).
For \( a_{10} \), \( n = 10 \):
\( a_{10} = -12 + (10-1) \times 5 \)
\( = -12 + 9 \times 5 \)
\( = -12 + 45 \)
\( = 33 \)
The first term is -12 and the 10th term is 33.
In simple words: We know the total of the first 9 numbers and how much each number changes. Using a special formula, we first find the very first number in the pattern. Then, using that first number and the change amount, we find what the 10th number in the pattern would be.

🎯 Exam Tip: Remember that the sum formula \( S_n \) helps you find the first term if other values are known, and the nth term formula \( a_n \) helps you find any specific term. Always calculate the first term accurately as subsequent calculations depend on it.

Question 6. The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and thirteenth term of the AP.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
Given that the sum of the first six terms \( S_6 = 42 \).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_6 = \frac{6}{2}[2a + (6-1)d] \)
\( 42 = 3[2a + 5d] \)
\( 42 = 6a + 15d \) ... (i)
Given that the ratio of the 10th term to the 30th term is 1 : 3.
The formula for the nth term is \( a_n = a + (n-1)d \).
So, \( a_{10} = a + (10-1)d = a + 9d \).
And \( a_{30} = a + (30-1)d = a + 29d \).
The ratio is \( \frac{a+9d}{a+29d} = \frac{1}{3} \).
Cross-multiply:
\( 3(a + 9d) = 1(a + 29d) \)
\( 3a + 27d = a + 29d \)
\( 3a - a = 29d - 27d \)
\( 2a = 2d \)
\( \implies a = d \)
Now, substitute \( a = d \) into equation (i):
\( 6a + 15a = 42 \)
\( 21a = 42 \)
\( \implies a = \frac{42}{21} \)
\( a = 2 \)
Since \( a = d \), we have \( d = 2 \).
So, the first term is 2.
Next, we need to calculate the thirteenth term, \( a_{13} \).
\( a_{13} = a + (13-1)d \)
\( = 2 + 12 \times 2 \)
\( = 2 + 24 \)
\( = 26 \)
In simple words: We used two pieces of information given about the number pattern: the sum of the first six numbers and the way the 10th and 30th numbers compare. By setting up equations from these facts, we first found that the starting number and the common difference are the same. Then, we solved for their actual value and used that to find the 13th number in the pattern.

🎯 Exam Tip: When working with ratios of terms, express each term using the \( a_n = a + (n-1)d \) formula. This helps set up an equation relating 'a' and 'd'. Always solve for 'a' and 'd' accurately before calculating any specific terms or sums.

Question 7. The 13th term of an A.P. is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
Given that the 13th term is four times its 3rd term:
\( a_{13} = 4 \times a_3 \)
Using the formula \( a_n = a + (n-1)d \):
\( a + 12d = 4(a + 2d) \)
\( a + 12d = 4a + 8d \)
\( 12d - 8d = 4a - a \)
\( 4d = 3a \)
\( \implies a = \frac{4}{3}d \) ... (Equation 1)
Also given that the fifth term is 16:
\( a_5 = 16 \)
Using the formula for the nth term:
\( a + (5-1)d = 16 \)
\( a + 4d = 16 \) ... (Equation 2)
Now, substitute the value of \( a \) from Equation 1 into Equation 2:
\( \frac{4}{3}d + 4d = 16 \)
To add the fractions, find a common denominator:
\( \frac{4d + 12d}{3} = 16 \)
\( \frac{16d}{3} = 16 \)
Multiply both sides by 3 and divide by 16:
\( d = \frac{16 \times 3}{16} \)
\( d = 3 \)
Now that we have \( d \), substitute it back into Equation 1 to find \( a \):
\( a = \frac{4}{3} \times 3 \)
\( a = 4 \)
So, the first term is 4 and the common difference is 3.
Finally, we need to find the sum of the first ten terms \( S_{10} \).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_{10} = \frac{10}{2}[2 \times 4 + (10-1) \times 3] \)
\( = 5[8 + 9 \times 3] \)
\( = 5[8 + 27] \)
\( = 5[35] \)
\( = 175 \)
In simple words: We used the given rules about the 13th and 3rd terms to find a link between the first term and the common difference. Then, we used the value of the 5th term to find the exact first term and common difference. With these, we calculated the total sum of the first ten numbers in the sequence. Arithmetic progressions often involve solving simultaneous equations to find 'a' and 'd'.

🎯 Exam Tip: When given multiple conditions about an A.P., translate each condition into an algebraic equation using the \( a_n = a + (n-1)d \) or \( S_n = \frac{n}{2}[2a+(n-1)d] \) formulas. Solve these equations simultaneously to find 'a' and 'd', which are crucial for finding other terms or sums.

Question 8. Find the 60th term of the AP 8,10,12,... if it has a total of 60 terms. Hence find the sum of its last 10 terms.
Answer:
For the given A.P: 8, 10, 12, ...
The first term \( a = 8 \).
The common difference \( d = 10 - 8 = 2 \).
The total number of terms is 60, so \( N = 60 \).
First, find the 60th term, \( T_{60} \). Using the formula \( T_n = a + (n-1)d \):
\( T_{60} = 8 + (60-1) \times 2 \)
\( = 8 + 59 \times 2 \)
\( = 8 + 118 \)
\( = 126 \)
Next, find the sum of its last 10 terms. The last 10 terms are from the 51st term to the 60th term.
The sum of the last 10 terms can be found by subtracting the sum of the first 50 terms from the sum of all 60 terms: \( S_{60} - S_{50} \).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_{60} = \frac{60}{2}[2 \times 8 + (60-1) \times 2] \)
\( = 30[16 + 59 \times 2] \)
\( = 30[16 + 118] \)
\( = 30[134] \)
\( = 4020 \)
\( S_{50} = \frac{50}{2}[2 \times 8 + (50-1) \times 2] \)
\( = 25[16 + 49 \times 2] \)
\( = 25[16 + 98] \)
\( = 25[114] \)
\( = 2850 \)
Sum of last 10 terms \( = S_{60} - S_{50} = 4020 - 2850 = 1170 \).
Alternatively, we can calculate this as:
\( S_{60} - S_{50} = \frac{60}{2}(2a+59d) - \frac{50}{2}(2a+49d) \)
\( = 30(2a+59d) - 25(2a+49d) \)
\( = 60a + 1770d - 50a - 1225d \)
\( = 10a + 545d \)
Substitute \( a=8 \) and \( d=2 \):
\( = 10(8) + 545(2) \)
\( = 80 + 1090 \)
\( = 1170 \)
In simple words: First, we found the 60th number in the pattern using the start number and the regular increase. Then, to find the sum of just the last 10 numbers, we calculated the total sum of all 60 numbers and subtracted the total sum of the first 50 numbers. This leaves us with the sum of the last 10 numbers alone.

🎯 Exam Tip: To find the sum of a specific block of terms within an A.P. (e.g., last 10 terms, terms from 20th to 30th), use the strategy of subtracting sums: \( S_{\text{end}} - S_{\text{start}-1} \). This is generally more efficient and less error-prone than calculating the first term of the sub-sequence and then its sum.

Question 9. In an A.P. if the 12th term is -13 and the sum of its first four terms is 24, find the sum of its first 10 terms.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
Given that the 12th term \( a_{12} = -13 \).
Using the formula \( a_n = a + (n-1)d \):
\( a + (12-1)d = -13 \)
\( a + 11d = -13 \) ... (Equation 1)
Given that the sum of its first four terms \( S_4 = 24 \).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_4 = \frac{4}{2}[2a + (4-1)d] \)
\( 24 = 2[2a + 3d] \)
Divide both sides by 2:
\( 12 = 2a + 3d \) ... (Equation 2)
Now we have a system of two linear equations with two variables \( a \) and \( d \).
From Equation 1, express \( a \) in terms of \( d \):
\( a = -13 - 11d \)
Substitute this expression for \( a \) into Equation 2:
\( 12 = 2(-13 - 11d) + 3d \)
\( 12 = -26 - 22d + 3d \)
\( 12 + 26 = -19d \)
\( 38 = -19d \)
\( \implies d = \frac{38}{-19} \)
\( d = -2 \)
Now, substitute the value of \( d \) back into the expression for \( a \):
\( a = -13 - 11(-2) \)
\( = -13 + 22 \)
\( = 9 \)
So, the first term is 9 and the common difference is -2.
Finally, we need to find the sum of the first ten terms \( S_{10} \).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( S_{10} = \frac{10}{2}[2 \times 9 + (10-1)(-2)] \)
\( = 5[18 + 9(-2)] \)
\( = 5[18 - 18] \)
\( = 5[0] \)
\( = 0 \)
In simple words: We used the given facts about the 12th term and the sum of the first four terms to create two equations. Solving these equations together helped us find the starting number and how much the numbers change. With those two key pieces of information, we then calculated the total sum of the first ten numbers in the pattern.

🎯 Exam Tip: Always be careful with negative numbers, especially when substituting values. Solving simultaneous equations accurately is key to finding 'a' and 'd', which are essential for further calculations in arithmetic progressions. A sum of zero often occurs when positive and negative terms cancel each other out in a decreasing A.P.

Question 10. Find the sum of the natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer:
Numbers that are divisible by both 2 and 5 are divisible by their least common multiple, which is 10.
We are looking for natural numbers between 101 and 999 that are divisible by 10.
The first number greater than 101 divisible by 10 is 110.
The last number less than 999 divisible by 10 is 990.
So, our Arithmetic Progression (A.P.) is 110, 120, 130, ..., 990.
Here, the first term \( a = 110 \).
The common difference \( d = 10 \).
The last term \( l = 990 \).
First, find the number of terms \( n \) using the formula \( l = a + (n-1)d \):
\( 990 = 110 + (n-1) \times 10 \)
\( 990 - 110 = 10(n-1) \)
\( 880 = 10(n-1) \)
\( \implies \frac{880}{10} = n-1 \)
\( 88 = n-1 \)
\( n = 88 + 1 \)
\( n = 89 \)
Now, find the sum of these 89 terms using the formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{89} = \frac{89}{2}[110 + 990] \)
\( = \frac{89}{2}[1100] \)
\( = 89 \times 550 \)
\( = 48950 \)
In simple words: We want to add up all the numbers between 101 and 999 that can be divided perfectly by both 2 and 5 (which means they can be divided by 10). We list these numbers, find the first and last ones, and count how many there are. Then, we use a quick sum formula to add them all up.

🎯 Exam Tip: The phrase "between X and Y" usually means excluding X and Y themselves. When numbers need to be divisible by multiple factors, find their Least Common Multiple (LCM) first. Then, identify the smallest and largest terms of the A.P. within the given range.

Question 11. Find the two digits numbers greatest than 50 which when divided by 7, leave a remainder of 4.
Answer:
We are looking for two-digit numbers greater than 50 that leave a remainder of 4 when divided by 7.
Let's list numbers that leave a remainder of 4 when divided by 7:
If \( k \) is an integer, the numbers are of the form \( 7k + 4 \).
For two-digit numbers greater than 50:
If \( k=7 \), \( 7 \times 7 + 4 = 49 + 4 = 53 \). This is the first such number greater than 50.
If \( k=8 \), \( 7 \times 8 + 4 = 56 + 4 = 60 \).
If \( k=9 \), \( 7 \times 9 + 4 = 63 + 4 = 67 \).
If \( k=10 \), \( 7 \times 10 + 4 = 70 + 4 = 74 \).
If \( k=11 \), \( 7 \times 11 + 4 = 77 + 4 = 81 \).
If \( k=12 \), \( 7 \times 12 + 4 = 84 + 4 = 88 \).
If \( k=13 \), \( 7 \times 13 + 4 = 91 + 4 = 95 \). This is the last two-digit number.
If \( k=14 \), \( 7 \times 14 + 4 = 98 + 4 = 102 \), which is a three-digit number.
So, the two-digit numbers greater than 50 which, when divided by 7, leave a remainder of 4 are: 53, 60, 67, 74, 81, 88, 95.
This sequence forms an A.P. with first term \( a = 53 \), common difference \( d = 7 \), and last term \( l = 95 \).
We can verify the count of these numbers using \( l = a + (n-1)d \):
\( 95 = 53 + (n-1)7 \)
\( 95 - 53 = 7(n-1) \)
\( 42 = 7(n-1) \)
\( \implies \frac{42}{7} = n-1 \)
\( 6 = n-1 \)
\( n = 6 + 1 \)
\( n = 7 \)
There are 7 such numbers. If asked for their sum:
\( S_7 = \frac{7}{2}[2 \times 53 + (7-1) \times 7] \)
\( = \frac{7}{2}[106 + 6 \times 7] \)
\( = \frac{7}{2}[106 + 42] \)
\( = \frac{7}{2}[148] \)
\( = 7 \times 74 \)
\( = 518 \)
In simple words: We are looking for numbers that are between 50 and 99 (two digits) and when you divide them by 7, you always have 4 left over. We find the first such number by adding 4 to a multiple of 7 that is just above 50. Then we keep adding 7 to find the next numbers until we reach the end of the two-digit range. This way, we list all the numbers that fit the rule.

🎯 Exam Tip: To find numbers satisfying "leaves a remainder of R when divided by D", write them in the form \( Dk + R \). Then, test values of \( k \) to find the first and last terms within the specified range. Listing the terms clearly helps to avoid mistakes.

Question 12. Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9.
Answer:
(i) Sum of integers between 100 and 200 that are divisible by 9:
The integers between 100 and 200 are 101, 102, ..., 199.
The first integer in this range divisible by 9 is 108 (since \( 9 \times 12 = 108 \)).
The last integer in this range divisible by 9 is 198 (since \( 9 \times 22 = 198 \)).
So, the A.P. is 108, 117, 126, ..., 198.
Here, the first term \( a = 108 \).
The common difference \( d = 9 \).
The last term \( l = 198 \).
First, find the number of terms \( n \) using \( l = a + (n-1)d \):
\( 198 = 108 + (n-1)9 \)
\( 198 - 108 = 9(n-1) \)
\( 90 = 9(n-1) \)
\( \implies \frac{90}{9} = n-1 \)
\( 10 = n-1 \)
\( n = 10 + 1 \)
\( n = 11 \)
Now, find the sum of these 11 terms using \( S_n = \frac{n}{2}(a+l) \):
\( S_{11} = \frac{11}{2}[108 + 198] \)
\( = \frac{11}{2}[306] \)
\( = 11 \times 153 \)
\( = 1683 \)

(ii) Sum of integers between 100 and 200 that are not divisible by 9:
First, find the sum of all integers between 100 and 200 (i.e., from 101 to 199).
This forms an A.P. with \( a = 101 \), \( d = 1 \), and \( l = 199 \).
Number of terms \( n = 199 - 101 + 1 = 99 \).
Sum of all integers \( S_{total} = \frac{n}{2}(a+l) \):
\( S_{total} = \frac{99}{2}[101 + 199] \)
\( = \frac{99}{2}[300] \)
\( = 99 \times 150 \)
\( = 14850 \)
The sum of integers not divisible by 9 is the total sum of integers minus the sum of integers divisible by 9.
Sum not divisible by 9 \( = S_{total} - S_{divisible by 9} \)
\( = 14850 - 1683 \)
\( = 13167 \)
In simple words: First, we find all the numbers between 100 and 200 that divide perfectly by 9, then we add them up. Second, we find the total sum of all numbers between 100 and 200. Finally, to get the sum of numbers that are NOT divisible by 9, we subtract the first sum from the total sum.

🎯 Exam Tip: When dealing with "not divisible by", it's often easiest to find the total sum of all numbers in the range and subtract the sum of numbers that *are* divisible by the given factor. Ensure the range (inclusive/exclusive) is handled carefully for "between" questions.

Question 13. Find the middle term of the sequence formed by all the numbers between 9 and 95 which leave a remainder 1 when divided by 3. Also find the sum of numbers on both the sides of the middle term separately.
Answer:
We need to find numbers between 9 and 95 that leave a remainder of 1 when divided by 3.
Numbers are of the form \( 3k + 1 \).
For numbers between 9 and 95:
If \( k=3 \), \( 3 \times 3 + 1 = 10 \). (The first number greater than 9)
If \( k=31 \), \( 3 \times 31 + 1 = 93 + 1 = 94 \). (The last number less than 95)
The sequence is 10, 13, 16, ..., 94.
This is an A.P. with first term \( a = 10 \), common difference \( d = 3 \), and last term \( l = 94 \).
Find the number of terms \( n \) using \( l = a + (n-1)d \):
\( 94 = 10 + (n-1)3 \)
\( 94 - 10 = 3(n-1) \)
\( 84 = 3(n-1) \)
\( \implies \frac{84}{3} = n-1 \)
\( 28 = n-1 \)
\( n = 28 + 1 \)
\( n = 29 \)
Since there are 29 terms, the middle term is the \( \frac{29+1}{2} = \frac{30}{2} = 15 \)th term.
Value of the middle term (\( a_{15} \)):
\( a_{15} = a + (15-1)d \)
\( = 10 + 14 \times 3 \)
\( = 10 + 42 \)
\( = 52 \)
(Note: The source calculates \( 10 + 15 \times 3 = 55 \), which corresponds to \( a + 15d \) rather than \( a + (15-1)d \). We will follow the established definition for \( a_n \).)

Sum of numbers on both sides of the middle term separately:
Since \( n=29 \) and the middle term is the 15th term, there are 14 terms before it and 14 terms after it.
Sum of the first 14 terms (before the middle term):
\( S_{14} = \frac{14}{2}[2a + (14-1)d] \)
\( = 7[2 \times 10 + 13 \times 3] \)
\( = 7[20 + 39] \)
\( = 7[59] \)
\( = 413 \)
Sum of the next 14 terms (after the middle term):
The first term of this sequence is \( a_{16} = a + 15d = 10 + 15 \times 3 = 10 + 45 = 55 \).
This is an A.P. with first term \( a' = 55 \), common difference \( d' = 3 \), and number of terms \( n' = 14 \).
\( S'_{14} = \frac{14}{2}[2a' + (14-1)d'] \)
\( = 7[2 \times 55 + 13 \times 3] \)
\( = 7[110 + 39] \)
\( = 7[149] \)
\( = 1043 \)
In simple words: First, we listed all the numbers between 9 and 95 that leave a remainder of 1 when divided by 3. Then, we counted these numbers to find out there are 29 of them. The middle number is the 15th one, which is 52. After that, we calculated the total of the first 14 numbers (before the middle one) and separately calculated the total of the next 14 numbers (after the middle one).

🎯 Exam Tip: When finding a middle term for an A.P. with an odd number of terms \( n \), the position is \( \frac{n+1}{2} \). Ensure you use the correct formula \( a_k = a + (k-1)d \) to find its value. For sums of terms on "either side," calculate the new first term for the second half of the sequence carefully.

Question 14. Sum of first n terms of an A.P., prove that \( S_{12} = 3(S_8 - S_4) \).
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the sum of the first \( n \) terms of an A.P. is \( S_n = \frac{n}{2}[2a+(n-1)d] \).

Let's calculate the Right Hand Side (RHS) of the equation: \( 3(S_8 - S_4) \).
First, find \( S_8 \):
\( S_8 = \frac{8}{2}[2a + (8-1)d] = 4[2a + 7d] \)
Next, find \( S_4 \):
\( S_4 = \frac{4}{2}[2a + (4-1)d] = 2[2a + 3d] \)
Now, calculate \( S_8 - S_4 \):
\( S_8 - S_4 = 4(2a + 7d) - 2(2a + 3d) \)
\( = (8a + 28d) - (4a + 6d) \)
\( = 8a + 28d - 4a - 6d \)
\( = 4a + 22d \)
Now, multiply by 3 for the RHS:
\( \text{RHS} = 3(4a + 22d) = 12a + 66d \)

Now, let's calculate the Left Hand Side (LHS) of the equation: \( S_{12} \).
Using the sum formula for \( n=12 \):
\( S_{12} = \frac{12}{2}[2a + (12-1)d] \)
\( = 6[2a + 11d] \)
\( = 12a + 66d \)

Since LHS \( = 12a + 66d \) and RHS \( = 12a + 66d \), we have LHS = RHS.
Therefore, \( S_{12} = 3(S_8 - S_4) \) is proven.
In simple words: To prove this, we first wrote out the formulas for the sums of 4, 8, and 12 terms using the general sum rule for number patterns. Then, we calculated both sides of the given equation separately. We showed that both sides simplify to the exact same expression, which means the statement is true.

🎯 Exam Tip: For "prove that" questions, always simplify both the Left Hand Side (LHS) and the Right Hand Side (RHS) independently until they match. Clearly state the formulas you are using and show all algebraic steps. Make sure to use \( S_n \) correctly for each value of \( n \).

Question 15. Show that the sum of first n even natural numbers is equal to \( (1 + \frac { 1 }{ n }) \) times the sum of the first n odd natural numbers.
Answer:
Let's find the sum of the first \( n \) even natural numbers.
The sequence of first \( n \) even natural numbers is 2, 4, 6, ..., \( 2n \).
This is an A.P. with first term \( a = 2 \), common difference \( d = 2 \), and \( n \) terms.
Sum of first \( n \) even natural numbers \( (S_{\text{even}}) \):
\( S_{\text{even}} = \frac{n}{2}[2a + (n-1)d] \)
\( = \frac{n}{2}[2(2) + (n-1)2] \)
\( = \frac{n}{2}[4 + 2n - 2] \)
\( = \frac{n}{2}[2 + 2n] \)
\( = \frac{n}{2} \times 2(1 + n) \)
\( = n(1 + n) \)

Next, let's find the sum of the first \( n \) odd natural numbers.
The sequence of first \( n \) odd natural numbers is 1, 3, 5, ..., \( (2n-1) \).
This is an A.P. with first term \( a = 1 \), common difference \( d = 2 \), and \( n \) terms.
Sum of first \( n \) odd natural numbers \( (S_{\text{odd}}) \):
\( S_{\text{odd}} = \frac{n}{2}[2a + (n-1)d] \)
\( = \frac{n}{2}[2(1) + (n-1)2] \)
\( = \frac{n}{2}[2 + 2n - 2] \)
\( = \frac{n}{2}[2n] \)
\( = n^2 \)

Now, we need to show that \( S_{\text{even}} = (1 + \frac { 1 }{ n }) S_{\text{odd}} \).
Let's evaluate the Right Hand Side (RHS) of the required equality:
\( \text{RHS} = (1 + \frac { 1 }{ n }) S_{\text{odd}} \)
\( = \left(\frac{n+1}{n}\right) \times n^2 \)
\( = (n+1) \times n \)
\( = n(n+1) \)
Since \( S_{\text{even}} = n(n+1) \) and RHS \( = n(n+1) \), we have LHS = RHS.
Therefore, the sum of first \( n \) even natural numbers is equal to \( (1 + \frac { 1 }{ n }) \) times the sum of the first \( n \) odd natural numbers is proven.
In simple words: We first found a simple formula for the total sum of the first 'n' even numbers and another simple formula for the total sum of the first 'n' odd numbers. Then, we used these formulas to check if the given statement holds true by calculating both sides of the equation. Both sides gave the same answer, so the statement is correct.

🎯 Exam Tip: Clearly define the A.P. for even and odd numbers (first term, common difference, number of terms). Remember that the sum of the first n odd numbers is always \( n^2 \), and the sum of the first n even numbers is \( n(n+1) \). These are common results worth remembering or deriving quickly.

Question 16. The sum of n terms of an A.P. whose first term is 5 and common difference is 36 is equal to the sum of 2n terms of another A.P. whose first term is 36 and common difference is 5. Find n.
Answer:
Let's consider the first A.P.:
First term \( a_1 = 5 \).
Common difference \( d_1 = 36 \).
Number of terms \( = n \).
The sum of \( n \) terms for the first A.P. is \( S_{n,1} = \frac{n}{2}[2a_1 + (n-1)d_1] \).
\( S_{n,1} = \frac{n}{2}[2(5) + (n-1)36] \)
\( = \frac{n}{2}[10 + 36n - 36] \)
\( = \frac{n}{2}[36n - 26] \)
\( = n(18n - 13) \)

Now, let's consider the second A.P.:
First term \( a_2 = 36 \).
Common difference \( d_2 = 5 \).
Number of terms \( = 2n \).
The sum of \( 2n \) terms for the second A.P. is \( S_{2n,2} = \frac{2n}{2}[2a_2 + (2n-1)d_2] \).
\( S_{2n,2} = n[2(36) + (2n-1)5] \)
\( = n[72 + 10n - 5] \)
\( = n[67 + 10n] \)

According to the problem statement, the sum of \( n \) terms of the first A.P. is equal to the sum of \( 2n \) terms of the second A.P.:
\( S_{n,1} = S_{2n,2} \)
\( n(18n - 13) = n(67 + 10n) \)
Since \( n \) represents the number of terms, \( n \) cannot be zero. So, we can divide both sides by \( n \):
\( 18n - 13 = 67 + 10n \)
Group the terms with \( n \) on one side and constant terms on the other:
\( 18n - 10n = 67 + 13 \)
\( 8n = 80 \)
\( \implies n = \frac{80}{8} \)
\( n = 10 \)
In simple words: We set up equations for the sum of terms for two different number patterns based on the information given. One pattern had 'n' terms, and the other had '2n' terms. Since their sums were equal, we put the two sum equations together and solved them to find the value of 'n'.

🎯 Exam Tip: When dealing with two separate A.P.s, clearly label their first terms and common differences (e.g., \( a_1, d_1 \) and \( a_2, d_2 \)) to avoid confusion. Remember that if \( n \) represents the number of terms, it must be a positive integer, allowing you to divide by \( n \) in an equation if it appears on both sides.

Question 17. If the sum of first ten terms of an A.P. is 4 times the sum of the first five terms, then find the ratio of the first term of the common difference.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the sum of the first \( n \) terms of an A.P. is \( S_n = \frac{n}{2}[2a+(n-1)d] \).
Given that the sum of the first ten terms is 4 times the sum of the first five terms: \( S_{10} = 4 \times S_5 \).

First, find \( S_{10} \):
\( S_{10} = \frac{10}{2}[2a + (10-1)d] \)
\( = 5[2a + 9d] \)

Next, find \( S_5 \):
\( S_5 = \frac{5}{2}[2a + (5-1)d] \)
\( = \frac{5}{2}[2a + 4d] \)

Now, substitute these into the given condition \( S_{10} = 4 \times S_5 \):
\( 5(2a + 9d) = 4 \times \frac{5}{2}(2a + 4d) \)
\( 5(2a + 9d) = 10(2a + 4d) \)
Divide both sides by 5 to simplify:
\( 2a + 9d = 2(2a + 4d) \)
\( 2a + 9d = 4a + 8d \)
Group the terms with \( a \) on one side and terms with \( d \) on the other:
\( 9d - 8d = 4a - 2a \)
\( d = 2a \)
We need to find the ratio of the first term to the common difference, i.e., \( \frac{a}{d} \).
From \( d = 2a \), we can write:
\( \frac{a}{d} = \frac{1}{2} \)
The ratio of the first term to the common difference is 1:2.
In simple words: We used the given relationship between the sum of the first ten numbers and the sum of the first five numbers in the pattern. By writing out the sum formulas and solving the resulting equation, we found a direct link between the starting number and the amount each number changes by. This link gave us the required ratio.

🎯 Exam Tip: Always write down the given conditions mathematically. Simplify expressions by cancelling common factors on both sides of an equation early on to make calculations easier. The final ratio should be presented in its simplest form, like \( \frac{1}{2} \) or 1:2.

Question 18. In a nursery, 37 plants have been arranged in the first row, 35 in the second, 33 in the third and so on. If there are 5 plants in the last row, how many plants are there in the nursery.
Answer:
The number of plants in each row forms an Arithmetic Progression (A.P.): 37, 35, 33, ...
Here, the first term \( a = 37 \).
The common difference \( d = 35 - 37 = -2 \).
The number of plants in the last row (which is the nth term) \( l = 5 \).
First, find the number of rows \( n \) using the formula for the nth term: \( l = a + (n-1)d \).
\( 5 = 37 + (n-1)(-2) \)
\( 5 - 37 = -2(n-1) \)
\( -32 = -2(n-1) \)
\( \implies \frac{-32}{-2} = n-1 \)
\( 16 = n-1 \)
\( n = 16 + 1 \)
\( n = 17 \)
So, there are 17 rows in the nursery.
Now, find the total number of plants in the nursery, which is the sum of all plants in all rows. Use the sum formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{17} = \frac{17}{2}[37 + 5] \)
\( = \frac{17}{2}[42] \)
\( = 17 \times 21 \)
\( = 357 \)
Thus, there are 357 plants in total in the nursery.
In simple words: The number of plants in each row forms a decreasing pattern. We first found out how many rows there are by using the number of plants in the first and last rows, and how much the number of plants decreases by in each row. Once we knew the total number of rows, we used a sum formula to add up all the plants to find the total in the nursery.

🎯 Exam Tip: This is a practical application of A.P. Be sure to correctly identify the first term, common difference, and the last term. If the problem asks for a total count, it usually implies finding the sum of all terms, but you might need to find the number of terms ('n') first.

Question 19. 200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next, above it, 18 in the row above it and so on. In how many rows are the 200 logs placed and how many logs are there in the top row?
Answer:
The number of logs in each row forms an Arithmetic Progression (A.P.): 20, 19, 18, ...
Here, the first term \( a = 20 \).
The common difference \( d = 19 - 20 = -1 \).
The total number of logs \( S_n = 200 \).
We need to find the number of rows \( n \) and the number of logs in the top row (\( a_n \)).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( 200 = \frac{n}{2}[2 \times 20 + (n-1)(-1)] \)
\( 200 = \frac{n}{2}[40 - n + 1] \)
\( 200 = \frac{n}{2}[41 - n] \)
Multiply both sides by 2:
\( 400 = n(41 - n) \)
\( 400 = 41n - n^2 \)
Rearrange into a quadratic equation:
\( n^2 - 41n + 400 = 0 \)
Factorize the quadratic equation. We need two numbers that multiply to 400 and add up to -41. These numbers are -16 and -25.
\( n^2 - 25n - 16n + 400 = 0 \)
Factor by grouping:
\( n(n - 25) - 16(n - 25) = 0 \)
\( (n - 25)(n - 16) = 0 \)
This gives two possible values for \( n \):
\( n - 25 = 0 \implies n = 25 \)
or
\( n - 16 = 0 \implies n = 16 \)

We must check both values for \( n \). The number of logs in any row cannot be negative.
Case 1: If \( n = 25 \)
The number of logs in the 25th row (\( a_{25} \)) would be:
\( a_{25} = a + (25-1)d \)
\( = 20 + 24(-1) \)
\( = 20 - 24 \)
\( = -4 \)
Since the number of logs cannot be negative, \( n=25 \) is not a possible answer.

Case 2: If \( n = 16 \)
The number of logs in the 16th row (\( a_{16} \)) would be:
\( a_{16} = a + (16-1)d \)
\( = 20 + 15(-1) \)
\( = 20 - 15 \)
\( = 5 \)
This is a valid number of logs.
Therefore, the number of rows is 16, and there are 5 logs in the top row.
In simple words: We used the starting number of logs and how they decrease in each row, along with the total number of logs, to find out how many rows there are. We solved a puzzle-like equation that gave two possible answers for the number of rows. After checking, only one of these answers made sense (you can't have negative logs!), which helped us find both the total number of rows and how many logs were in the very last row.

🎯 Exam Tip: When a quadratic equation yields two possible values for 'n', always verify both by calculating the corresponding \( a_n \). Discard any value of 'n' that results in a physically impossible outcome, such as a negative number of items or terms. The context of the problem is crucial for valid solutions.

Question 20. A sum of Rs. 1890 is to be used to give seven cash prizes. If each prize is Rs. 50 less than the preceding prize, find the value of each prize?
Answer:
Let the first prize be Rs. \( a \).
Since each prize is Rs. 50 less than the preceding prize, the common difference \( d = -50 \).
The total number of prizes \( n = 7 \).
The total sum of the prizes \( S_n = \text{Rs. } 1890 \).
Using the sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \):
\( 1890 = \frac{7}{2}[2a + (7-1)(-50)] \)
\( 1890 = \frac{7}{2}[2a + 6(-50)] \)
\( 1890 = \frac{7}{2}[2a - 300] \)
Multiply both sides by 2 and divide by 7:
\( \frac{1890 \times 2}{7} = 2a - 300 \)
\( 270 \times 2 = 2a - 300 \)
\( 540 = 2a - 300 \)
\( 540 + 300 = 2a \)
\( 840 = 2a \)
\( \implies a = \frac{840}{2} \)
\( a = 420 \)
So, the first prize is Rs. 420.
Now we can find the value of each of the seven prizes:
First prize \( (a_1) = \text{Rs. } 420 \)
Second prize \( (a_2) = a + d = 420 - 50 = \text{Rs. } 370 \)
Third prize \( (a_3) = 370 - 50 = \text{Rs. } 320 \)
Fourth prize \( (a_4) = 320 - 50 = \text{Rs. } 270 \)
Fifth prize \( (a_5) = 270 - 50 = \text{Rs. } 220 \)
Sixth prize \( (a_6) = 220 - 50 = \text{Rs. } 170 \)
Seventh prize \( (a_7) = 170 - 50 = \text{Rs. } 120 \)
The values of the seven cash prizes are Rs. 420, Rs. 370, Rs. 320, Rs. 270, Rs. 220, Rs. 170, and Rs. 120.
In simple words: We know the total money for prizes, the number of prizes, and that each prize is Rs. 50 less than the one before it. We used a special formula to find the value of the very first prize. Once we knew the first prize, we subtracted Rs. 50 repeatedly to find the value of each of the seven prizes.

🎯 Exam Tip: Pay close attention to keywords like "less than the preceding prize," which indicates a negative common difference. After finding the first term, list all terms explicitly as requested, ensuring each value decreases by the common difference.

Question 20. A sum of Rs 1890 is to be used to give seven cash prizes. If each prize is Rs 50 less than the preceding prize, find the value of each prize?
Answer: The total amount for the cash prizes is Rs 1890. There are seven cash prizes in total. Each prize is Rs 50 less than the one before it, so the common difference of this arithmetic progression is \( d = -50 \). Let 'a' be the value of the first prize. We use the formula for the sum of an Arithmetic Progression (AP): \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Now, we substitute the known values into the formula:
\( 1890 = \frac{7}{2}[2a + (7-1)(-50)] \)
\( 1890 = \frac{7}{2}[2a + 6(-50)] \)
\( 1890 = \frac{7}{2}[2a - 300] \)
To find 'a', we multiply both sides by 2 and then divide by 7:
\( \frac{1890 \times 2}{7} = 2a - 300 \)
\( 540 = 2a - 300 \)
Add 300 to both sides to isolate the term with 'a':
\( 540 + 300 = 2a \)
\( 840 = 2a \)
Finally, divide by 2 to find 'a':
\( a = \frac{840}{2} \)
\( a = 420 \)
The first prize is Rs 420. To find the values of all seven prizes, we subtract Rs 50 consecutively. The seven cash prizes are:
Rs 420
Rs \( 420 - 50 = 370 \)
Rs \( 370 - 50 = 320 \)
Rs \( 320 - 50 = 270 \)
Rs \( 270 - 50 = 220 \)
Rs \( 220 - 50 = 170 \)
Rs \( 170 - 50 = 120 \)
These prizes form a decreasing arithmetic progression.
In simple words: We used the total prize money, the number of prizes, and the rule that each prize is Rs 50 less than the last one. Using a special math formula for sums, we found the first prize was Rs 420. Then, we just subtracted Rs 50 each time to get the value for all seven prizes.

🎯 Exam Tip: When solving problems involving arithmetic progressions for prizes or similar decreasing values, remember to use a negative common difference (d) to show the reduction in value.