---
title: "NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1"
url: https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-1
dateModified: 2026-10-07T15:51:43+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

Chapter 5: Arithmetic Progressions. Every question from Exercise 5.1, with full working and the final answer.

Free PDF (29 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-c1e752f614.pdf

## EXERCISE 5.1

### Question 1

*3 marks · Short answer*

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.

**Part (i)**

1. The fare for the first km is ₹ 15 and ₹ 8 for each additional km.
2. The list of fares for 1 km, 2 km, 3 km, 4 km, etc. is 15, 23, 31, 39, ...
3. Since the difference between consecutive terms is constant ($23 - 15 = 8$, $31 - 23 = 8$), it forms an AP.

Answer (i): It forms an AP because each term increases by a fixed amount of ₹ 8.

**Part (ii)**

1. Let the initial amount of air in the cylinder be $V$.
2. When the pump removes $\frac{1}{4}$ of the remaining air, the amounts of air left successively are $V, \frac{3}{4}V, \frac{3}{4}\left(\frac{3}{4}V\right) = \frac{9}{16}V, \dots$
3. The differences between consecutive terms are not the same, so it does not form an AP.

Answer (ii): It does not form an AP as the successive differences are not equal.

**Part (iii)**

1. The cost of digging for the first metre is ₹ 150 and rises by ₹ 50 for each subsequent metre.
2. The costs for 1 m, 2 m, 3 m, 4 m, etc. are 150, 200, 250, 300, ...
3. Since the difference between consecutive terms is constant ($200 - 150 = 50$, $250 - 200 = 50$), it forms an AP.

Answer (iii): It forms an AP because each term increases by a fixed amount of ₹ 50.

**Part (iv)**

1. The initial principal is ₹ 10000 deposited at compound interest at 8% per annum.
2. The amounts at the end of every year form a geometric progression, not an arithmetic progression.
3. The difference between successive amounts is not constant.

Answer (iv): It does not form an AP because the amounts grow by compound interest, yielding unequal differences.

**Answer:** Situations (i) and (iii) form an AP, whereas (ii) and (iv) do not.

> Common mistake: Confusing compound interest growth with arithmetic progression.

### Question 2

*3 marks · Short answer*

Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:
(i) $a = 10, d = 10$
(ii) $a = -2, d = 0$
(iii) $a = 4, d = -3$
(iv) $a = -1, d = \frac{1}{2}$
(v) $a = -1.25, d = -0.25$

**Part (i)**

1. Given $a = 10$ and $d = 10$.
2. The first four terms are $10, 10+10, 10+2(10), 10+3(10)$.

Answer (i): $10, 20, 30, 40$

**Part (ii)**

1. Given $a = -2$ and $d = 0$.
2. The first four terms are $-2, -2+0, -2+2(0), -2+3(0)$.

Answer (ii): $-2, -2, -2, -2$

**Part (iii)**

1. Given $a = 4$ and $d = -3$.
2. The first four terms are $4, 4+(-3), 4+2(-3), 4+3(-3)$.

Answer (iii): $4, 1, -2, -5$

**Part (iv)**

1. Given $a = -1$ and $d = \frac{1}{2}$.
2. The first four terms are $-1, -1+\frac{1}{2}, -1+2\left(\frac{1}{2}\right), -1+3\left(\frac{1}{2}\right)$.

Answer (iv): $-1, -\frac{1}{2}, 0, \frac{1}{2}$

**Part (v)**

1. Given $a = -1.25$ and $d = -0.25$.
2. The first four terms are $-1.25, -1.25+(-0.25), -1.25+2(-0.25), -1.25+3(-0.25)$.

Answer (v): $-1.25, -1.50, -1.75, -2.00$

**Answer:** The first four terms for the given values are computed using $a, a+d, a+2d, a+3d$.

> Common mistake: Arithmetic errors when adding negative numbers or fractions.

### Question 3

*3 marks · Short answer*

For the following APs, write the first term and the common difference:
(i) $3, 1, -1, -3, \dots$
(ii) $-5, -1, 3, 7, \dots$
(iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
(iv) $0.6, 1.7, 2.8, 3.9, \dots$

**Part (i)**

1. The given AP is $3, 1, -1, -3, \dots$
2. The first term $a = 3$.
3. The common difference $d = 1 - 3 = -2$.

Answer (i): $a = 3, d = -2$

**Part (ii)**

1. The given AP is $-5, -1, 3, 7, \dots$
2. The first term $a = -5$.
3. The common difference $d = -1 - (-5) = 4$.

Answer (ii): $a = -5, d = 4$

**Part (iii)**

1. The given AP is $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
2. The first term $a = \frac{1}{3}$.
3. The common difference $d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$.

Answer (iii): $a = \frac{1}{3}, d = \frac{4}{3}$

**Part (iv)**

1. The given AP is $0.6, 1.7, 2.8, 3.9, \dots$
2. The first term $a = 0.6$.
3. The common difference $d = 1.7 - 0.6 = 1.1$.

Answer (iv): $a = 0.6, d = 1.1$

**Answer:** First term and common difference obtained for each AP.

> Common mistake: Subtracting in the wrong order when finding $d$, such as $a_1 - a_2$ instead of $a_2 - a_1$.

### Question 4

*5 marks · Long answer*

Which of the following are APs ? If they form an AP, find the common difference $d$ and write three more terms.
(i) $2, 4, 8, 16, \dots$
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
(iii) $-1.2, -3.2, -5.2, -7.2, \dots$
(iv) $-10, -6, -2, 2, \dots$
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
(vi) $0.2, 0.22, 0.222, 0.2222, \dots$
(vii) $0, -4, -8, -12, \dots$
(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
(ix) $1, 3, 9, 27, \dots$
(x) $a, 2a, 3a, 4a, \dots$
(xi) $a, a^2, a^3, a^4, \dots$
(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$
(xiv) $1^2, 3^2, 5^2, 7^2, \dots$
(xv) $1^2, 5^2, 7^2, 73, \dots$

**Part (i)**

1. Given list: $2, 4, 8, 16, \dots$
2. Here $a_2 - a_1 = 4 - 2 = 2$ and $a_3 - a_2 = 8 - 4 = 4$.
3. Since $a_2 - a_1 \neq a_3 - a_2$, it is not an AP.

Answer (i): Not an AP.

**Part (ii)**

1. Given list: $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
2. Here $a_2 - a_1 = \frac{5}{2} - 2 = \frac{1}{2}$ and $a_3 - a_2 = 3 - \frac{5}{2} = \frac{1}{2}$.
3. It is an AP with $d = \frac{1}{2}$. The next three terms are $4, \frac{9}{2}, 5$.

Answer (ii): Is an AP; $d = \frac{1}{2}$; next three terms: $4, \frac{9}{2}, 5$

**Part (iii)**

1. Given list: $-1.2, -3.2, -5.2, -7.2, \dots$
2. Here $a_2 - a_1 = -3.2 - (-1.2) = -2$ and $a_3 - a_2 = -5.2 - (-3.2) = -2$.
3. It is an AP with $d = -2$. The next three terms are $-9.2, -11.2, -13.2$.

Answer (iii): Is an AP; $d = -2$; next three terms: $-9.2, -11.2, -13.2$

**Part (iv)**

1. Given list: $-10, -6, -2, 2, \dots$
2. Here $a_2 - a_1 = -6 - (-10) = 4$ and $a_3 - a_2 = -2 - (-6) = 4$.
3. It is an AP with $d = 4$. The next three terms are $6, 10, 14$.

Answer (iv): Is an AP; $d = 4$; next three terms: $6, 10, 14$

**Part (v)**

1. Given list: $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
2. Here $a_2 - a_1 = \sqrt{2}$ and $a_3 - a_2 = \sqrt{2}$.
3. It is an AP with $d = \sqrt{2}$. The next three terms are $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$.

Answer (v): Is an AP; $d = \sqrt{2}$; next three terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$

**Part (vi)**

1. Given list: $0.2, 0.22, 0.222, 0.2222, \dots$
2. Here $a_2 - a_1 = 0.02$ and $a_3 - a_2 = 0.002$.
3. Since $a_2 - a_1 \neq a_3 - a_2$, it is not an AP.

Answer (vi): Not an AP.

**Part (vii)**

1. Given list: $0, -4, -8, -12, \dots$
2. Here $a_2 - a_1 = -4$ and $a_3 - a_2 = -4$.
3. It is an AP with $d = -4$. The next three terms are $-16, -20, -24$.

Answer (vii): Is an AP; $d = -4$; next three terms: $-16, -20, -24$

**Part (viii)**

1. Given list: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
2. Here $a_2 - a_1 = 0$ and $a_3 - a_2 = 0$.
3. It is an AP with $d = 0$. The next three terms are $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.

Answer (viii): Is an AP; $d = 0$; next three terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$

**Part (ix)**

1. Given list: $1, 3, 9, 27, \dots$
2. Here $a_2 - a_1 = 2$ and $a_3 - a_2 = 6$.
3. Since differences are not equal, it is not an AP.

Answer (ix): Not an AP.

**Part (x)**

1. Given list: $a, 2a, 3a, 4a, \dots$
2. Here $a_2 - a_1 = a$ and $a_3 - a_2 = a$.
3. It is an AP with $d = a$. The next three terms are $5a, 6a, 7a$.

Answer (x): Is an AP; $d = a$; next three terms: $5a, 6a, 7a$

**Part (xi)**

1. Given list: $a, a^2, a^3, a^4, \dots$
2. Here $a_2 - a_1 = a^2 - a$ and $a_3 - a_2 = a^3 - a^2$.
3. Since these differences are not equal, it is not an AP.

Answer (xi): Not an AP.

**Part (xii)**

1. Given list: $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$, which simplifies to $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$
2. Here $a_2 - a_1 = \sqrt{2}$ and $a_3 - a_2 = \sqrt{2}$.
3. It is an AP with $d = \sqrt{2}$. The next three terms are $5\sqrt{2} (=\sqrt{50}), 6\sqrt{2} (=\sqrt{72}), 7\sqrt{2} (=\sqrt{98})$.

Answer (xii): Is an AP; $d = \sqrt{2}$; next three terms: $\sqrt{50}, \sqrt{72}, \sqrt{98}$

**Part (xiii)**

1. Given list: $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$
2. Here $a_2 - a_1 = \sqrt{6} - \sqrt{3}$ and $a_3 - a_2 = 3 - \sqrt{6}$.
3. Since these differences are not equal, it is not an AP.

Answer (xiii): Not an AP.

**Part (xiv)**

1. Given list: $1^2, 3^2, 5^2, 7^2, \dots$, i.e., $1, 9, 25, 49, \dots$
2. Here $a_2 - a_1 = 8$ and $a_3 - a_2 = 16$.
3. Since differences are not equal, it is not an AP.

Answer (xiv): Not an AP.

**Part (xv)**

1. Given list: $1^2, 5^2, 7^2, 73, \dots$, i.e., $1, 25, 49, 73, \dots$
2. Here $a_2 - a_1 = 24$ and $a_3 - a_2 = 24$ and $a_4 - a_3 = 24$.
3. It is an AP with $d = 24$. The next three terms are $97, 121, 145$.

Answer (xv): Is an AP; $d = 24$; next three terms: $97, 121, 145$

**Answer:** Checked all lists for AP and wrote common differences and next three terms where applicable.

> Common mistake: Assuming lists like $\sqrt{2}, \sqrt{8}, \sqrt{18}$ are not APs without simplifying the surds first.

## Related pages

- [All Chapter 5 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions)
- [Exercise 5.2](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-2)
- [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)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student