---
title: "NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2"
url: https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-2
dateModified: 2026-10-07T15:51:43+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2

Chapter 5: Arithmetic Progressions. Every question from Exercise 5.2, with full working and the final answer.

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## EXERCISE 5.2

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP:
(i) $a = 7, d = 3, n = 8, a_n = \dots$
(ii) $a = -18, d = \dots, n = 10, a_n = 0$
(iii) $a = \dots, d = -3, n = 18, a_n = -5$
(iv) $a = -18.9, d = 2.5, n = \dots, a_n = 3.6$
(v) $a = 3.5, d = 0, n = 105, a_n = \dots$

**Part (i)**

1. Using $a_n = a + (n-1)d$, substitute $a = 7$, $d = 3$, and $n = 8$.
2. $a_8 = 7 + (8-1) \times 3 = 7 + 21 = 28$.

Answer (i): 28

**Part (ii)**

1. Using $a_n = a + (n-1)d$, substitute $a = -18$, $n = 10$, and $a_{10} = 0$.
2. $0 = -18 + (10-1)d \implies 9d = 18 \implies d = 2$.

Answer (ii): 2

**Part (iii)**

1. Using $a_n = a + (n-1)d$, substitute $d = -3$, $n = 18$, and $a_{18} = -5$.
2. $-5 = a + (18-1)(-3) \implies -5 = a - 51 \implies a = 46$.

Answer (iii): 46

**Part (iv)**

1. Using $a_n = a + (n-1)d$, substitute $a = -18.9$, $d = 2.5$, and $a_n = 3.6$.
2. $3.6 = -18.9 + (n-1) \times 2.5 \implies 22.5 = (n-1) \times 2.5 \implies n-1 = 9 \implies n = 10$.

Answer (iv): 10

**Part (v)**

1. Using $a_n = a + (n-1)d$, substitute $a = 3.5$, $d = 0$, and $n = 105$.
2. $a_{105} = 3.5 + (105-1) \times 0 = 3.5$.

Answer (v): 3.5

**Answer:** (i) 28, (ii) 2, (iii) 46, (iv) 10, (v) 3.5

> Common mistake: Arithmetic errors while transposing negative signs in the linear equation for $n$ or $d$.

### Question 2

*1 mark · MCQ*

Choose the correct choice in the following and justify :
(i) 30th term of the AP: $10, 7, 4, \dots$, is
(ii) 11th term of the AP: $-3, -\frac{1}{2}, 2, \dots$, is

- 97, 77, -77, -87
- 28, 22, -38, -48\frac{1}{2}

**Part (i)**

1. Here $a = 10$, $d = 7 - 10 = -3$, and $n = 30$.
2. $a_{30} = 10 + (30-1)(-3) = 10 + 29(-3) = 10 - 87 = -77$.

Answer (i): (C) -77

**Part (ii)**

1. Here $a = -3$, $d = -\frac{1}{2} - (-3) = 2.5 = \frac{5}{2}$, and $n = 11$.
2. $a_{11} = -3 + (11-1)\left(\frac{5}{2}\right) = -3 + 10 \times \frac{5}{2} = -3 + 25 = 22$.

Answer (ii): (B) 22

**Answer:** (i) (C) -77, (ii) (B) 22

> Common mistake: Incorrectly calculating the common difference by subtracting the first term from the second incorrectly.

### Question 3

*3 marks · Short answer*

In the following APs, find the missing terms in the boxes :
(i) $2, \Box, 26$
(ii) $\Box, 13, \Box, 3$
(iii) $5, \Box, \Box, 9\frac{1}{2}$
(iv) $-4, \Box, \Box, \Box, \Box, 6$
(v) $\Box, 38, \Box, \Box, \Box, -22$

**Part (i)**

1. Let the terms be $a_1, a_2, a_3$. Here $a_1 = 2$ and $a_3 = 26$.
2. We know $a_3 = a_1 + 2d$, so $26 = 2 + 2d$, which gives $d = 12$.
3. The missing term is $a_2 = a_1 + d = 2 + 12 = 14$.

Answer (i): 14

**Part (ii)**

1. Here $a_2 = 13$ and $a_4 = 3$. We have $a_2 = a + d = 13$ and $a_4 = a + 3d = 3$.
2. Subtracting the first equation from the second gives $2d = -10$, so $d = -5$.
3. Then $a = 18$ and the missing terms are $a_1 = 18$ and $a_3 = 8$.

Answer (ii): 18, 8

**Part (iii)**

1. Here $a_1 = 5$ and $a_4 = 9\frac{1}{2} = \frac{19}{2}$.
2. We have $a_4 = a_1 + 3d$, so $\frac{19}{2} = 5 + 3d$, which gives $3d = \frac{9}{2}$, so $d = \frac{3}{2}$.
3. The missing terms are $a_2 = 5 + \frac{3}{2} = 6\frac{1}{2}$ and $a_3 = 6\frac{1}{2} + \frac{3}{2} = 8$.

Answer (iii): 6\frac{1}{2}, 8

**Part (iv)**

1. Here $a_1 = -4$ and $a_6 = 6$. We have $a_6 = a + 5d$, so $6 = -4 + 5d$, which gives $5d = 10$, so $d = 2$.
2. The missing terms are $a_2 = -2$, $a_3 = 0$, $a_4 = 2$, and $a_5 = 4$.

Answer (iv): -2, 0, 2, 4

**Part (v)**

1. Here $a_2 = 38$ and $a_6 = -22$. We have $a+d=38$ and $a+5d=-22$.
2. Subtracting gives $4d = -60$, so $d = -15$.
3. Then $a = 53$, and the missing terms are $a_1 = 53$, $a_3 = 23$, $a_4 = 8$, and $a_5 = -7$.

Answer (v): 53, 23, 8, -7

**Answer:** The missing terms for each AP are given in the respective parts.

> Common mistake: Mixing up the formula for $a_n$ leading to incorrect common difference $d$.

### Question 4

*3 marks · Short answer*

Which term of the AP : $3, 8, 13, 18, \dots$, is 78?

**Solution**

1. Given AP is $3, 8, 13, 18, \dots$, so first term $a = 3$ and common difference $d = 8 - 3 = 5$.
2. Let 78 be the $n$th term of the given AP, so $a_n = 78$.
3. Using the formula $a_n = a + (n-1)d$, we get $3 + (n-1)5 = 78$.
4. Solving for $n$, we get $(n-1)5 = 75 \implies n-1 = 15 \implies n = 16$.
5. Therefore, the 16th term of the AP is 78.

**Answer:** 16th term

> Common mistake: Mixing up $a_n$ and $n$ in the formula.

### Question 5

*3 marks · Short answer*

Find the number of terms in each of the following APs :
(i) $7, 13, 19, \dots, 205$
(ii) $18, 15\frac{1}{2}, 13, \dots, -47$

**Part (i)**

1. Given AP: $7, 13, 19, \dots, 205$. Here $a = 7$, $d = 6$, and $a_n = 205$.
2. Using $a_n = a + (n-1)d$, we have $7 + (n-1)6 = 205$.
3. $6(n-1) = 198 \implies n-1 = 33 \implies n = 34$.

Answer (i): 34

**Part (ii)**

1. Given AP: $18, 15\frac{1}{2}, 13, \dots, -47$. Here $a = 18$, $d = -2.5 = -\frac{5}{2}$, and $a_n = -47$.
2. Using $a_n = a + (n-1)d$, we have $18 + (n-1)\left(-\frac{5}{2}\right) = -47$.
3. $(n-1)\left(-\frac{5}{2}\right) = -65 \implies n-1 = 26 \implies n = 27$.

Answer (ii): 27

**Answer:** (i) 34, (ii) 27

> Common mistake: Calculation errors with fractional common differences.

### Question 6

*3 marks · Short answer*

Check whether $-150$ is a term of the AP : $11, 8, 5, 2 \dots$

**Solution**

1. Given AP is $11, 8, 5, 2, \dots$, so first term $a = 11$ and common difference $d = 8 - 11 = -3$.
2. Let $-150$ be the $n$th term of the given AP, so $a_n = -150$.
3. Using the formula $a_n = a + (n-1)d$, we get $11 + (n-1)(-3) = -150$.
4. Simplifying, we get $-3(n-1) = -161 \implies n-1 = \frac{161}{3} \implies n = \frac{164}{3}$.
5. Since $n$ is not a positive integer, $-150$ is not a term of the given AP.

**Answer:** -150 is not a term of the given AP

> Common mistake: Concluding that a non-integer $n$ can be a valid index.

### Question 7

*3 marks · Short answer*

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

**Solution**

1. Let the first term be $a$ and the common difference be $d$.
2. We are given $a_{11} = 38$, so $a + 10d = 38$.
3. We are also given $a_{16} = 73$, so $a + 15d = 73$.
4. Subtracting the first equation from the second gives $5d = 35$, so $d = 7$.
5. Substituting $d = 7$ in $a + 10d = 38$ gives $a + 70 = 38$, so $a = -32$.
6. Now, find the 31st term: $a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178$.

**Answer:** 178

> Common mistake: Arithmetic error while solving simultaneous equations or calculating $a_{31}$.

### Question 8

*3 marks · Short answer*

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

**Solution**

1. An AP consists of 50 terms, so $n = 50$, and the last term is $a_{50} = 106$.
2. We are given the 3rd term $a_3 = 12$, so $a + 2d = 12$.
3. The last term is $a_{50} = 106$, so $a + 49d = 106$.
4. Subtracting the two equations gives $47d = 94$, so $d = 2$.
5. Substituting $d = 2$ in $a + 2d = 12$ gives $a + 4 = 12$, so $a = 8$.
6. Now, find the 29th term: $a_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$.

**Answer:** 64

> Common mistake: Taking the last term as $a_{n}$ instead of $a_{50}$.

### Question 9

*3 marks · Short answer*

If the 3rd and the 9th terms of an AP are 4 and $-8$ respectively, which term of this AP is zero?

**Solution**

1. We are given $a_3 = 4$, so $a + 2d = 4$.
2. We are given $a_9 = -8$, so $a + 8d = -8$.
3. Subtracting the first equation from the second gives $6d = -12$, so $d = -2$.
4. Substituting $d = -2$ in $a + 2d = 4$ gives $a - 4 = 4$, so $a = 8$.
5. Let the $n$th term of the AP be zero, so $a_n = a + (n - 1)d = 0$.
6. Substituting the values, $8 + (n - 1)(-2) = 0$, which gives $2(n - 1) = 8$, so $n - 1 = 4$, and $n = 5$.

**Answer:** 5th term

> Common mistake: Sign errors while solving for $d$ and $n$.

### Question 10

*3 marks · Short answer*

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

**Solution**

1. The 17th term of an AP is given by $a_{17} = a + 16d$.
2. The 10th term of the AP is given by $a_{10} = a + 9d$.
3. We are given that $a_{17} - a_{10} = 7$.
4. Substitute the expressions: $(a + 16d) - (a + 9d) = 7$.
5. Simplify the equation: $7d = 7$.
6. Therefore, the common difference $d = 1$.

**Answer:** 1

> Common mistake: Including the first term $a$ in the final answer when it gets cancelled out.

### Question 11

*3 marks · Short answer*

Which term of the AP : $3, 15, 27, 39, \dots$ will be 132 more than its 54th term?

**Solution**

1. For the given AP $3, 15, 27, 39, \dots$, we have first term $a = 3$ and common difference $d = 12$.
2. The 54th term is given by $a_{54} = a + 53d = 3 + 53(12) = 3 + 636 = 639$.
3. We need to find the term which is $132$ more than $a_{54}$, so let this be $a_n$.
4. Therefore, $a_n = 639 + 132 = 771$.
5. Using the formula $a_n = a + (n - 1)d$, we have $3 + (n - 1)12 = 771$.
6. Solving this gives $(n - 1)12 = 768$, so $n - 1 = 64$, and $n = 65$.

**Answer:** 65th term

> Common mistake: Calculation error in multiplying $53 \times 12$.

### Question 12

*3 marks · Short answer*

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

**Solution**

1. Let the first term of the first AP be $a$ and the first term of the second AP be $A$, with both having the same common difference $d$.
2. The 100th term of the first AP is $a_{100} = a + 99d$.
3. The 100th term of the second AP is $A_{100} = A + 99d$.
4. We are given that the difference between their 100th terms is $100$, so $(a + 99d) - (A + 99d) = 100$, which gives $a - A = 100$.
5. Now, find the difference between their 1000th terms: $(a_{1000} - A_{1000}) = (a + 999d) - (A + 999d) = a - A$.
6. Since $a - A = 100$, the difference between their 1000th terms is also $100$.

**Answer:** 100

> Common mistake: Assuming the difference changes with the index of the term.

### Question 13

*3 marks · Short answer*

How many three-digit numbers are divisible by 7?

**Solution**

1. The list of three-digit numbers divisible by 7 is 105, 112, 119, ..., 994.
2. This forms an AP with first term $a = 105$, common difference $d = 7$, and last term $a_n = 994$.
3. Using the formula $a_n = a + (n - 1)d$, we have $994 = 105 + (n - 1) \times 7$.
4. $889 = (n - 1) \times 7$, which gives $n - 1 = 127$.
5. Thus, $n = 128$, so there are 128 three-digit numbers divisible by 7.

**Answer:** 128

> Common mistake: Taking the first three-digit number divisible by 7 incorrectly as 100 or 102.

### Question 14

*3 marks · Short answer*

How many multiples of 4 lie between 10 and 250?

**Solution**

1. The multiples of 4 that lie between 10 and 250 are 12, 16, 20, ..., 248.
2. This forms an AP with first term $a = 12$, common difference $d = 4$, and last term $a_n = 248$.
3. Using the formula $a_n = a + (n - 1)d$, we have $248 = 12 + (n - 1) \times 4$.
4. $236 = (n - 1) \times 4$, which gives $n - 1 = 59$.
5. Thus, $n = 60$, so there are 60 multiples of 4 between 10 and 250.

**Answer:** 60

> Common mistake: Including 8 or 252 in the list.

### Question 15

*3 marks · Short answer*

For what value of $n$, are the $n$th terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

**Solution**

1. For the first AP: $63, 65, 67, \dots$, first term $a_1 = 63$ and common difference $d_1 = 2$.
2. Its $n$th term is given by $a_n = 63 + (n - 1) \times 2 = 61 + 2n$.
3. For the second AP: $3, 10, 17, \dots$, first term $a_2 = 3$ and common difference $d_2 = 7$.
4. Its $n$th term is given by $a_n = 3 + (n - 1) \times 7 = 7n - 4$.
5. Equating the two $n$th terms: $61 + 2n = 7n - 4$, which gives $5n = 65$, so $n = 13$.

**Answer:** 13

> Common mistake: Algebraic error while simplifying $63 + 2(n-1)$.

### Question 16

*3 marks · Short answer*

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

**Solution**

1. We are given that the third term is 16, so $a_3 = a + 2d = 16$.
2. The 7th term exceeds the 5th term by 12, so $a_7 - a_5 = 12$, which means $(a + 6d) - (a + 4d) = 12$.
3. Simplifying this gives $2d = 12$, so $d = 6$.
4. Substituting $d = 6$ into $a + 2d = 16$, we get $a + 12 = 16$, so $a = 4$.
5. Thus, the required AP is $4, 10, 16, 22, \dots$.

**Answer:** $4, 10, 16, 22, \dots$

> Common mistake: Writing $a_7 - a_5$ incorrectly without expanding using the general term formula.

### Question 17

*3 marks · Short answer*

Find the 20th term from the last term of the AP : $3, 8, 13, \dots, 253$.

**Solution**

1. For the given AP: $3, 8, 13, \dots, 253$, first term $a = 3$, common difference $d = 5$, and last term $l = 253$.
2. To find the 20th term from the last term, we can reverse the AP where the first term becomes $253$ and the common difference becomes $-5$.
3. Using the formula $a_n = a + (n - 1)d$ with $a = 253$, $d = -5$, and $n = 20$, we get $a_{20} = 253 + (20 - 1)(-5)$.
4. $a_{20} = 253 + 19(-5) = 253 - 95 = 158$.
5. Therefore, the 20th term from the last term is 158.

**Answer:** 158

> Common mistake: Subtracting 20 times the common difference instead of 19 times.

### Question 18

*3 marks · Short answer*

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

**Solution**

1. We are given that $a_4 + a_8 = 24$, which can be written as $(a + 3d) + (a + 7d) = 24$, or $2a + 10d = 24$, simplifying to $a + 5d = 12$.
2. We are also given that $a_6 + a_{10} = 44$, which can be written as $(a + 5d) + (a + 9d) = 44$, or $2a + 14d = 44$, simplifying to $a + 7d = 22$.
3. Subtracting the first equation from the second gives $(a + 7d) - (a + 5d) = 22 - 12$, so $2d = 10$, which means $d = 5$.
4. Substituting $d = 5$ into $a + 5(5) = 12$, we get $a + 25 = 12$, so $a = -13$.
5. The first three terms are $a = -13$, $a + d = -13 + 5 = -8$, and $a + 2d = -13 + 10 = -3$, giving $-13, -8, -3$.

**Answer:** $-13, -8, -3$

> Common mistake: Sign errors while solving the linear equations for $a$ and $d$.

### Question 19

*3 marks · Short answer*

Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

**Solution**

1. The annual salaries for the years 1995, 1996, 1997, … form an AP with first term $a = 5000$ and common difference $d = 200$.
2. Let his income reach ₹ 7000 in the $n$th year, so $a_n = 7000$.
3. Using the formula $a_n = a + (n - 1)d$, we get $7000 = 5000 + (n - 1) \times 200$.
4. Solving this gives $2000 = (n - 1) \times 200$, which implies $n - 1 = 10$, so $n = 11$.
5. Therefore, his income reached ₹ 7000 in the 11th year, which corresponds to the year $1995 + 11 - 1 = 2005$.

**Answer:** 2005

> Common mistake: Confusing the number of terms $n$ with the calendar year.

### Question 20

*3 marks · Short answer*

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n$th week, her weekly savings become ₹ 20.75, find $n$.

**Solution**

1. The weekly savings form an AP with first term $a = 5$ and common difference $d = 1.75$.
2. Given that the $n$th term $a_n = 20.75$.
3. Using the formula $a_n = a + (n - 1)d$, we have $20.75 = 5 + (n - 1) \times 1.75$.
4. Subtracting 5 from both sides gives $15.75 = (n - 1) \times 1.75$.
5. Dividing by 1.75 gives $n - 1 = 9$, so $n = 10$.

**Answer:** 10

> Common mistake: Arithmetic errors while dividing decimals.

## Related pages

- [All Chapter 5 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions)
- [Exercise 5.1](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-1)
- [Exercise 5.3](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-3)
- [Exercise 5.4](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-4)

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