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

# NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

This chapter's questions cover the identification, nth term, and sum of terms of arithmetic progressions (APs), along with various word problems applied to real-life situations. It includes exercises and optional problems designed to test students' mastery of AP concepts.

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.

## 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.

## EXERCISE 5.3

### Question 1

*3 marks · Case-based*

Find the sum of the following APs:
(i) $2, 7, 12, \dots$, to 10 terms.
(ii) $-37, -33, -29, \dots$, to 12 terms.
(iii) $0.6, 1.7, 2.8, \dots$, to 100 terms.
(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to 11 terms.

**Part (i)**

1. Here, first term $a = 2$, common difference $d = 7 - 2 = 5$, and number of terms $n = 10$.
2. We use the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
3. Substituting the values, $S_{10} = \frac{10}{2}[2(2) + (10 - 1)5] = 5[4 + 45] = 5 \times 49 = 245$.

Answer (i): 245

**Part (ii)**

1. Here, first term $a = -37$, common difference $d = -33 - (-37) = 4$, and number of terms $n = 12$.
2. We use the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
3. Substituting the values, $S_{12} = \frac{12}{2}[2(-37) + (12 - 1)4] = 6[-74 + 44] = 6(-30) = -180$.

Answer (ii): -180

**Part (iii)**

1. Here, first term $a = 0.6$, common difference $d = 1.7 - 0.6 = 1.1$, and number of terms $n = 100$.
2. We use the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
3. Substituting the values, $S_{100} = \frac{100}{2}[2(0.6) + (100 - 1)1.1] = 50[1.2 + 99 \times 1.1] = 50[1.2 + 108.9] = 50 \times 110.1 = 5505$.

Answer (iii): 5505

**Part (iv)**

1. Here, first term $a = \frac{1}{15}$, common difference $d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$, and number of terms $n = 11$.
2. We use the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
3. Substituting the values, $S_{11} = \frac{11}{2}\left[2\left(\frac{1}{15}\right) + (11 - 1)\left(\frac{1}{60}\right)\right] = \frac{11}{2}\left(\frac{2}{15} + \frac{10}{60}\right) = \frac{11}{2}\left(\frac{2}{15} + \frac{1}{6}\right) = \frac{11}{2}\left(\frac{4 + 5}{30}\right) = \frac{11}{2} \times \frac{9}{30} = \frac{33}{20}$.

Answer (iv): \frac{33}{20}

**Answer:** The sums of the given APs are (i) 245, (ii) -180, (iii) 5505, (iv) 33/20.

> Common mistake: Arithmetic errors while simplifying fractions or decimals.

### Question 2

*3 marks · Short answer*

Find the sums given below :
(i) $7 + 10\frac{1}{2} + 14 + \dots + 84$
(ii) $34 + 32 + 30 + \dots + 10$
(iii) $-5 + (-8) + (-11) + \dots + (-230)$

**Part (i)**

1. Here $a = 7, d = 10\frac{1}{2} - 7 = 3.5 = \frac{7}{2}$, and $l = 84$.
2. Using $l = a + (n - 1)d$, we have $84 = 7 + (n - 1)\frac{7}{2}$, giving $77 = (n - 1)\frac{7}{2}$, so $n - 1 = 22$, which means $n = 23$.
3. Using $S_n = \frac{n}{2}(a + l)$, we get $S_{23} = \frac{23}{2}(7 + 84) = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046\frac{1}{2}$.

Answer (i): $1046\frac{1}{2}$

**Part (ii)**

1. Here $a = 34, d = 32 - 34 = -2$, and $l = 10$.
2. Using $l = a + (n - 1)d$, we have $10 = 34 + (n - 1)(-2)$, giving $-24 = (n - 1)(-2)$, so $n - 1 = 12$, which means $n = 13$.
3. Using $S_n = \frac{n}{2}(a + l)$, we get $S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2}(44) = 13 \times 22 = 286$.

Answer (ii): 286

**Part (iii)**

1. Here $a = -5, d = -8 - (-5) = -3$, and $l = -230$.
2. Using $l = a + (n - 1)d$, we have $-230 = -5 + (n - 1)(-3)$, giving $-225 = (n - 1)(-3)$, so $n - 1 = 75$, which means $n = 76$.
3. Using $S_n = \frac{n}{2}(a + l)$, we get $S_{76} = \frac{76}{2}(-5 + (-230)) = 38(-235) = -8930$.

Answer (iii): -8930

**Answer:** (i) $1046\frac{1}{2}$, (ii) 286, (iii) -8930

> Common mistake: Always find the number of terms $n$ using the $n$-th term formula before applying the sum formula.

### Question 3

*5 marks · Case-based*

In an AP:
(i) given $a = 5, d = 3, a_n = 50$, find $n$ and $S_n$.
(ii) given $a = 7, a_{13} = 35$, find $d$ and $S_{13}$.
(iii) given $a_{12} = 37, d = 3$, find $a$ and $S_{12}$.
(iv) given $a_3 = 15, S_{10} = 125$, find $d$ and $a_{10}$.
(v) given $d = 5, S_9 = 75$, find $a$ and $a_9$.
(vi) given $a = 2, d = 8, S_n = 90$, find $n$ and $a_n$.
(vii) given $a = 8, a_n = 62, S_n = 210$, find $n$ and $d$.
(viii) given $a_n = 4, d = 2, S_n = -14$, find $n$ and $a$.
(ix) given $a = 3, n = 8, S = 192$, find $d$.
(x) given $l = 28, S = 144$, and there are total 9 terms. Find $a$.

**Part (i)**

1. Given $a = 5, d = 3, a_n = 50$. Using $a_n = a + (n - 1)d$, we get $50 = 5 + (n - 1)3$.
2. Solving for $n$, $45 = 3(n - 1) \implies n - 1 = 15 \implies n = 16$.
3. Now, $S_n = \frac{n}{2}(a + a_n)$ gives $S_{16} = \frac{16}{2}(5 + 50) = 8 \times 55 = 440$.

Answer (i): n = 16, S_{16} = 440

**Part (ii)**

1. Given $a = 7, a_{13} = 35$. Using $a_n = a + (n - 1)d$, we get $35 = 7 + (13 - 1)d$.
2. Solving for $d$, $28 = 12d \implies d = \frac{28}{12} = \frac{7}{3}$.
3. Now, $S_{13} = \frac{13}{2}(a + a_{13}) = \frac{13}{2}(7 + 35) = \frac{13}{2} \times 42 = 13 \times 21 = 273$.

Answer (ii): d = \frac{7}{3}, S_{13} = 273

**Part (iii)**

1. Given $a_{12} = 37, d = 3$. Using $a_{12} = a + 11d$, we get $37 = a + 11(3)$.
2. Solving for $a$, $37 = a + 33 \implies a = 4$.
3. Now, $S_{12} = \frac{12}{2}(a + a_{12}) = 6(4 + 37) = 6 \times 41 = 246$.

Answer (iii): a = 4, S_{12} = 246

**Part (iv)**

1. Given $a_3 = 15, S_{10} = 125$. We have $a + 2d = 15$ and $\frac{10}{2}[2a + 9d] = 125$.
2. Simplifying the second equation, $5(2a + 9d) = 125 \implies 2a + 9d = 25$.
3. From the first equation, $a = 15 - 2d$. Substituting this, $2(15 - 2d) + 9d = 25 \implies 30 - 4d + 9d = 25 \implies 5d = -5 \implies d = -1$.
4. Then $a = 15 - 2(-1) = 17$. Now, $a_{10} = a + 9d = 17 + 9(-1) = 8$.

Answer (iv): d = -1, a_{10} = 8

**Part (v)**

1. Given $d = 5, S_9 = 75$. Using $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $75 = \frac{9}{2}[2a + (9 - 1)5]$.
2. Simplifying, $75 = \frac{9}{2}[2a + 40] = 9(a + 20) \implies 25 = 3(a + 20) \implies 3a + 60 = 25 \implies 3a = -35 \implies a = -\frac{35}{3}$.
3. Now, $a_9 = a + 8d = -\frac{35}{3} + 8(5) = -\frac{35}{3} + 40 = \frac{-35 + 120}{3} = \frac{85}{3}$.

Answer (v): a = -\frac{35}{3}, a_9 = \frac{85}{3}

**Part (vi)**

1. Given $a = 2, d = 8, S_n = 90$. Using $S_n = \frac{n}{2}[2a + (n - 1)d]$, we have $90 = \frac{n}{2}[2(2) + (n - 1)8]$.
2. Simplifying, $180 = n[4 + 8n - 8] = n[8n - 4] = 4n[2n - 1] \implies 45 = n(2n - 1) \implies 2n^2 - n - 45 = 0$.
3. Solving the quadratic equation, $2n^2 - 10n + 9n - 45 = 0 \implies 2n(n - 5) + 9(n - 5) = 0 \implies (2n + 9)(n - 5) = 0$.
4. Since $n$ must be a positive integer, $n = 5$. Then $a_5 = a + 4d = 2 + 4(8) = 34$.

Answer (vi): n = 5, a_5 = 34

**Part (vii)**

1. Given $a = 8, a_n = 62, S_n = 210$. Using $S_n = \frac{n}{2}(a + a_n)$, we get $210 = \frac{n}{2}(8 + 62)$.
2. Simplifying, $210 = \frac{n}{2}(70) = 35n \implies n = \frac{210}{35} = 6$.
3. Now, using $a_n = a + (n - 1)d$, we get $62 = 8 + (6 - 1)d \implies 54 = 5d \implies d = \frac{54}{5}$.

Answer (vii): n = 6, d = \frac{54}{5}

**Part (viii)**

1. Given $a_n = 4, d = 2, S_n = -14$. Using $a_n = a + (n - 1)d$, we get $4 = a + (n - 1)2 \implies a = 4 - 2(n - 1) = 6 - 2n$.
2. Using $S_n = \frac{n}{2}(a + a_n)$, we get $-14 = \frac{n}{2}(a + 4) \implies -28 = n(a + 4)$.
3. Substituting $a = 6 - 2n$, we get $-28 = n(6 - 2n + 4) = n(10 - 2n) = 10n - 2n^2 \implies 2n^2 - 10n - 28 = 0 \implies n^2 - 5n - 14 = 0$.
4. Factoring, $(n - 7)(n + 2) = 0 \implies n = 7$ (since $n$ cannot be negative). Then $a = 6 - 2(7) = -8$.

Answer (viii): n = 7, a = -8

**Part (ix)**

1. Given $a = 3, n = 8, S = 192$. Using $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $192 = \frac{8}{2}[2(3) + (8 - 1)d]$.
2. Simplifying, $192 = 4[6 + 7d] \implies 48 = 6 + 7d \implies 7d = 42 \implies d = 6$.

Answer (ix): d = 6

**Part (x)**

1. Given $l = 28, S = 144$ (where $S = S_9$), and total terms $n = 9$.
2. Using the formula $S_n = \frac{n}{2}(a + l)$, we get $144 = \frac{9}{2}(a + 28)$.
3. Solving for $a$, $144 \times \frac{2}{9} = a + 28 \implies 16 \times 2 = a + 28 \implies 32 = a + 28 \implies a = 4$.

Answer (x): a = 4

**Answer:** All requested parameters for the 10 parts of question 3 have been calculated.

> Common mistake: Mixing up formula variables or rejecting valid/invalid values of n incorrectly.

### Question 4

*3 marks · Short answer*

How many terms of the AP : $9, 17, 25, \dots$ must be taken to give a sum of 636?

**Solution**

1. Here, the AP is $9, 17, 25, \dots$, so $a = 9$, $d = 17 - 9 = 8$, and $S_n = 636$.
2. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we substitute the values: $636 = \frac{n}{2}[2(9) + (n - 1)8]$.
3. Simplify the expression inside the brackets: $636 = \frac{n}{2}[18 + 8n - 8] = \frac{n}{2}[10 + 8n] = n[4n + 5]$.
4. Expand to form a quadratic equation: $4n^2 + 5n - 636 = 0$.
5. Factorize the quadratic equation: $4n^2 - 48n + 53n - 636 = 0$, which gives $4n(n - 12) + 53(n - 12) = 0$, or $(4n + 53)(n - 12) = 0$.
6. Since $n$ must be a positive integer, $n = 12$ (rejecting $n = -\frac{53}{4}$).
7. Therefore, 12 terms of the AP must be taken to give a sum of 636.

**Answer:** 12

> Common mistake: Not rejecting the negative fractional value of $n$.

### Question 5

*3 marks · Short answer*

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

**Solution**

1. Given first term $a = 5$, last term $a_n = l = 45$, and sum $S_n = 400$.
2. Using the formula $S_n = \frac{n}{2}(a + l)$, we substitute the values to get $400 = \frac{n}{2}(5 + 45)$.
3. Simplifying gives $400 = 25n$, so $n = \frac{400}{25} = 16$.
4. Using the formula $a_n = a + (n - 1)d$, we substitute $n = 16$ to get $45 = 5 + (16 - 1)d$.
5. Solving for $d$ gives $15d = 40$, so $d = \frac{40}{15} = \frac{8}{3}$.

**Answer:** $n = 16, d = \frac{8}{3}$

> Common mistake: Incorrectly using the $S_n$ formula with $d$ instead of $l$ when $d$ is unknown.

### Question 6

*3 marks · Short answer*

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

**Solution**

1. Given first term $a = 17$, last term $a_n = l = 350$, and common difference $d = 9$.
2. Using the formula $a_n = a + (n - 1)d$, substitute the values to get $350 = 17 + (n - 1) \times 9$.
3. Simplifying gives $333 = (n - 1) \times 9$, so $n - 1 = 37$ and $n = 38$.
4. Using the formula $S_n = \frac{n}{2}(a + l)$, substitute $n = 38$ to get $S_{38} = \frac{38}{2}(17 + 350)$.
5. Calculating the result gives $S_{38} = 19 \times 367 = 6973$.

**Answer:** $n = 38, S_{38} = 6973$

> Common mistake: Confusing the index $n$ with the sum value during calculation.

### Question 7

*3 marks · Short answer*

Find the sum of first 22 terms of an AP in which $d = 7$ and 22nd term is 149.

**Solution**

1. Given $d = 7$ and $a_{22} = 149$, we need to find the sum of the first 22 terms ($S_{22}$).
2. Using the formula for the $n$th term, $a_{n} = a + (n - 1)d$, we have $a_{22} = a + (22 - 1) \times 7 = 149$.
3. This gives $a + 21 \times 7 = 149$, so $a + 147 = 149$, which means $a = 2$.
4. Using the formula for the sum of $n$ terms, $S_{n} = \frac{n}{2}(a + l)$, we find $S_{22} = \frac{22}{2}(2 + 149) = 11 \times 151 = 1661$.

**Answer:** 1661

> Common mistake: Using the wrong formula for sum when the first term is not substituted properly.

### Question 8

*3 marks · Short answer*

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

**Solution**

1. Given $a_2 = 14$ and $a_3 = 18$, we can find the common difference $d = a_3 - a_2 = 18 - 14 = 4$.
2. Since $a_2 = a + d$, we have $a + 4 = 14$, which gives $a = 10$.
3. Using the sum formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$, the sum of the first 51 terms is $S_{51} = \frac{51}{2}[2(10) + (51 - 1) \times 4]$.
4. Calculating this gives $S_{51} = \frac{51}{2}[20 + 200] = \frac{51}{2} \times 220 = 51 \times 110 = 5610$.

**Answer:** 5610

> Common mistake: Calculation error while multiplying 51 by 110.

### Question 9

*3 marks · Short answer*

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first $n$ terms.

**Solution**

1. Given $S_7 = 49$ and $S_{17} = 289$, we use the formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
2. For $n = 7$, $\frac{7}{2}[2a + 6d] = 49$, which simplifies to $2a + 6d = 14$, or $a + 3d = 7$.
3. For $n = 17$, $\frac{17}{2}[2a + 16d] = 289$, which simplifies to $2a + 16d = 34$, or $a + 8d = 17$.
4. Subtracting the first equation from the second gives $5d = 10$, so $d = 2$, and substituting $d = 2$ gives $a = 1$.
5. The sum of the first $n$ terms is $S_{n} = \frac{n}{2}[2(1) + (n - 1)2] = \frac{n}{2}[2 + 2n - 2] = n^2$.

**Answer:** n^2

> Common mistake: Wrong simplification of linear equations in a and d.

### Question 10

*3 marks · Short answer*

Show that $a_1, a_2, \dots, a_n, \dots$ form an AP where $a_n$ is defined as below :
(i) $a_n = 3 + 4n$
(ii) $a_n = 9 - 5n$
Also find the sum of the first 15 terms in each case.

**Part (i)**

1. Given $a_{n} = 3 + 4n$, substituting $n = 1, 2, 3$ gives $a_1 = 7$, $a_2 = 11$, $a_{3} = 15$.
2. Since $a_2 - a_1 = 4$ and $a_3 - a_2 = 4$, the list forms an AP with $a = 7$ and $d = 4$.
3. The sum of the first 15 terms is $S_{15} = \frac{15}{2}[2(7) + (15 - 1) \times 4] = \frac{15}{2}[14 + 56] = \frac{15}{2} \times 70 = 525$.

Answer (i): 525

**Part (ii)**

1. Given $a_{n} = 9 - 5n$, substituting $n = 1, 2, 3$ gives $a_1 = 4$, $a_2 = -1$, $a_3 = -6$.
2. Since $a_2 - a_1 = -5$ and $a_3 - a_2 = -5$, the list forms an AP with $a = 4$ and $d = -5$.
3. The sum of the first 15 terms is $S_{15} = \frac{15}{2}[2(4) + (15 - 1) \times (-5)] = \frac{15}{2}[8 - 70] = \frac{15}{2} \times (-62) = -465$.

Answer (ii): -465

**Answer:** (i) AP with sum 525, (ii) AP with sum -465

> Common mistake: Sign errors while calculating terms with negative common difference.

### Question 11

*3 marks · Short answer*

If the sum of the first $n$ terms of an AP is $4n - n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the $n$th terms.

**Solution**

1. Given $S_{n} = 4n - n^2$, the first term $S_1 = a_1 = 4(1) - (1)^2 = 3$.
2. The sum of the first two terms is $S_2 = 4(2) - (2)^2 = 8 - 4 = 4$.
3. The second term is $a_2 = S_2 - S_1 = 4 - 3 = 1$.
4. The common difference is $d = a_2 - a_1 = 1 - 3 = -2$.
5. The 3rd term is $a_3 = a + 2d = 3 + 2(-2) = -1$, the 10th term is $a_{10} = a + 9d = 3 + 9(-2) = -15$, and the $n$th term is $a_{n} = a + (n - 1)d = 3 + (n - 1)(-2) = 5 - 2n$.

**Answer:** a_1 = 3, S_2 = 4, a_2 = 1, a_3 = -1, a_{10} = -15, a_n = 5 - 2n

> Common mistake: Confusing $S_2$ with $a_2$.

### Question 12

*3 marks · Short answer*

Find the sum of the first 40 positive integers divisible by 6.

**Solution**

1. The positive integers divisible by 6 form an AP: $6, 12, 18, 24, \dots$.
2. Here, the first term $a = 6$, common difference $d = 6$, and we need the sum of the first 40 terms, so $n = 40$.
3. Using the sum formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$, we substitute the values to get $S_{40} = \frac{40}{2}[2(6) + (40 - 1) \times 6]$.
4. Calculating this gives $S_{40} = 20[12 + 39 \times 6] = 20[12 + 234] = 20 \times 246 = 4920$.

**Answer:** 4920

> Common mistake: Taking $a_n = 40$ instead of $n = 40$.

### Question 13

*3 marks · Short answer*

Find the sum of the first 15 multiples of 8.

**Solution**

1. The first 15 multiples of 8 are $8, 16, 24, \dots$
2. This forms an AP with first term $a = 8$, common difference $d = 8$, and number of terms $n = 15$.
3. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $S_{15} = \frac{15}{2}[2(8) + (15 - 1) \times 8]$.
4. $S_{15} = \frac{15}{2}[16 + 112] = \frac{15}{2} \times 128 = 15 \times 64 = 960$.

**Answer:** 960

> Common mistake: Taking incorrect first term or number of terms.

### Question 14

*3 marks · Short answer*

Find the sum of the odd numbers between 0 and 50.

**Solution**

1. The odd numbers between 0 and 50 are $1, 3, 5, \dots, 49$.
2. This forms an AP with $a = 1$, $d = 2$, and last term $a_n = 49$.
3. Using $a_n = a + (n - 1)d$, we find $n$: $49 = 1 + (n - 1) \times 2$, which gives $n = 25$.
4. Using $S_n = \frac{n}{2}(a + l)$, we get $S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2} \times 50 = 625$.

**Answer:** 625

> Common mistake: Wrongly identifying the number of terms n as 50 or 49/2.

### Question 15

*3 marks · Short answer*

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

**Solution**

1. The penalties for each day form an AP: $200, 250, 300, \dots$ for 30 days.
2. Here, first term $a = 200$, common difference $d = 50$, and number of terms $n = 30$.
3. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, the total penalty is $S_{30} = \frac{30}{2}[2(200) + (30 - 1) \times 50]$.
4. $S_{30} = 15[400 + 29 \times 50] = 15[400 + 1450] = 15 \times 1850 = 27750$.

**Answer:** ₹ 27750

> Common mistake: Calculation error in multiplying 29 by 50.

### Question 16

*3 marks · Short answer*

A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

**Solution**

1. Let the seven cash prizes form an AP where $S_7 = 700$ and $d = -20$.
2. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we substitute $n = 7$: $700 = \frac{7}{2}[2a + (7 - 1)(-20)]$.
3. $100 = \frac{1}{2}[2a - 120] \implies 200 = 2a - 120 \implies 2a = 320 \implies a = 160$.
4. The values of the seven prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, and ₹ 40.

**Answer:** ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40

> Common mistake: Taking d as positive 20 instead of -20.

### Question 17

*3 marks · Short answer*

In a school, students thought of planting 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 the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

**Solution**

1. The number of trees planted by the three sections of each class from Class I to XII are $3 \times 1, 3 \times 2, 3 \times 3, \dots, 3 \times 12$.
2. This is an arithmetic progression: $3, 6, 9, \dots, 36$ with $a = 3$, $d = 3$, and $n = 12$.
3. Using the sum formula $S_n = \frac{n}{2}(a + l)$, we get $S_{12} = \frac{12}{2}(3 + 36)$.
4. $S_{12} = 6 \times 39 = 234$.

**Answer:** 234 trees

> Common mistake: Forgetting to multiply by 3 for the three sections of each class.

### Question 18

*3 marks · Short answer*

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, $\dots$ as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \frac{22}{7}$)

**Solution**

1. The lengths of successive semicircles are $l_1 = \pi r_1 = \pi (0.5)$, $l_2 = \pi (1.0)$, $l_3 = \pi (1.5)$, $\dots$ for 13 semicircles.
2. This forms an AP: $0.5\pi, 1.0\pi, 1.5\pi, \dots$ where $a = 0.5\pi$, $d = 0.5\pi$, and $n = 13$.
3. Using $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $S_{13} = \frac{13}{2}[2(0.5\pi) + (13 - 1)(0.5\pi)]$.
4. $S_{13} = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2} \times 7\pi = \frac{13}{2} \times 7 \times \frac{22}{7} = 143 \text{ cm}$.

**Answer:** 143 cm

> Common mistake: Using diameter instead of radius for the length of a semicircle.

### Question 19

*3 marks · Short answer*

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?

**Solution**

1. The number of logs in each row from the bottom to top forms an AP: $20, 19, 18, \dots$
2. Here, the first term $a = 20$, the common difference $d = 19 - 20 = -1$, and the total sum of logs $S_n = 200$.
3. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $200 = \frac{n}{2}[2(20) + (n - 1)(-1)]$.
4. Simplifying the equation gives $400 = n[40 - n + 1]$, which expands to $n^2 - 41n + 400 = 0$.
5. Factoring the quadratic equation gives $(n - 16)(n - 25) = 0$, so $n = 16$ or $n = 25$.
6. If $n = 25$, the 25th term is $a_{25} = 20 + (25 - 1)(-1) = -4$, which is impossible since the number of logs cannot be negative, hence $n = 16$.
7. The number of logs in the top row is the 16th term: $a_{16} = 20 + (16 - 1)(-1) = 5$.
8. Thus, the logs are placed in 16 rows and there are 5 logs in the top row.

**Answer:** 16 rows and 5 logs in the top row

> Common mistake: Accepting both values of n without checking if the number of terms makes the last term negative.

### Question 20

*3 marks · Short answer*

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

**Solution**

1. The distance run to pick up the first potato and drop it back in the bucket is $2 \times 5 = 10~\text{m}$.
2. The distance run for the second potato is $2 \times (5 + 3) = 16~\text{m}$, and for the third potato is $2 \times (5 + 3 + 3) = 22~\text{m}$.
3. The total distances run for each potato form an AP: $10, 16, 22, \dots$ with first term $a = 10$ and common difference $d = 6$.
4. We need to find the total distance for 10 potatoes, which is the sum of the first 10 terms ($S_{10}$).
5. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we substitute $n = 10$, $a = 10$, and $d = 6$.
6. $S_{10} = \frac{10}{2}[2(10) + (10 - 1)(6)] = 5[20 + 54] = 5 \times 74 = 370~\text{m}$.

**Answer:** $370~\text{m}$

> Common mistake: Forgetting to multiply the distances by 2 to account for the return trip to the bucket.

## EXERCISE 5.4 (Optional)*

### Question 1

*3 marks · Short answer*

Which term of the AP : $121, 117, 113, \dots$, is its first negative term?

**Solution**

1. Given the AP: $121, 117, 113, \dots$, the first term $a = 121$ and common difference $d = 117 - 121 = -4$.
2. We need to find the first negative term, so we set $a_n < 0$, which gives $a + (n - 1)d < 0$.
3. Substituting the values, $121 + (n - 1)(-4) < 0$, which simplifies to $121 - 4n + 4 < 0$ or $125 < 4n$.
4. This means $4n > 125$, so $n > \frac{125}{4} = 31.25$.
5. Since $n$ must be a positive integer, the smallest integer greater than $31.25$ is $32$.
6. Therefore, the 32nd term is the first negative term.

**Answer:** 32nd term

> Common mistake: Taking $n$ as 31 instead of rounding up to the next integer 32.

### Question 2

*3 marks · Short answer*

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

**Solution**

1. Let the first term be $a$ and the common difference be $d$.
2. Given that the sum of the third and seventh terms is $6$, we have $(a + 2d) + (a + 6d) = 6$, which simplifies to $2a + 8d = 6$ or $a + 4d = 3$.
3. Given that their product is $8$, we have $(a + 2d)(a + 6d) = 8$.
4. From the first equation, $a = 3 - 4d$. Substituting this into the product equation gives $(3 - 4d + 2d)(3 - 4d + 6d) = 8$, or $(3 - 2d)(3 + 2d) = 8$.
5. This yields $9 - 4d^2 = 8$, so $4d^2 = 1$, giving $d = \frac{1}{2}$ or $d = -\frac{1}{2}$.
6. Case 1: When $d = \frac{1}{2}$, $a = 3 - 4(\frac{1}{2}) = 1$. The sum of the first 16 terms is $S_{16} = \frac{16}{2}[2(1) + (16 - 1)(\frac{1}{2})] = 8[2 + \frac{15}{2}] = 8(\frac{19}{2}) = 76$.
7. Case 2: When $d = -\frac{1}{2}$, $a = 3 - 4(-\frac{1}{2}) = 5$. The sum of the first 16 terms is $S_{16} = \frac{16}{2}[2(5) + (16 - 1)(-\frac{1}{2})] = 8[10 - \frac{15}{2}] = 8(\frac{5}{2}) = 20$.

**Answer:** 76 or 20

> Common mistake: Dropping one of the possible values for the common difference $d$.

### Question 3

*3 marks · Short answer*

A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs?

**Solution**

1. Given: distance between rungs is $25\text{ cm}$, bottom rung length is $45\text{ cm}$, top rung length is $25\text{ cm}$, and total height is $2\frac{1}{2}\,\text{m} = 250\text{ cm}$.
2. The number of rungs is given by $\text{Number of rungs} = \frac{250}{25} + 1 = 11$.
3. The lengths of the rungs form an AP with first term $a = 45$, last term $l = 25$, and number of terms $n = 11$.
4. The total length of the wood required is the sum of 11 terms of this AP: $S_{11} = \frac{n}{2}(a + l) = \frac{11}{2}(45 + 25) = \frac{11}{2} \times 70 = 385\text{ cm}$.

**Answer:** $385\text{ cm}$

> Common mistake: Forgetting to add 1 when calculating the number of rungs, leading to $10$ rungs instead of $11$.

### Question 4

*3 marks · Short answer*

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$.

**Solution**

1. The house numbers are $1, 2, 3, \dots, 49$, which form an AP with $a = 1$ and $d = 1$.
2. Let the house number be $x$. The sum of the house numbers preceding $x$ is $S_{x-1} = \frac{x-1}{2}[2(1) + (x - 1 - 1)(1)] = \frac{(x-1)x}{2}$.
3. The sum of the house numbers following $x$ is equal to the total sum of all 49 houses minus the sum of the first $x$ houses, which is $S_{49} - S_x$.
4. Given that $S_{x-1} = S_{49} - S_x$, we can rewrite this as $S_{x-1} + S_x = S_{49}$.
5. Using the formula for the sum of $n$ terms, $\frac{(x-1)x}{2} + \frac{x(x+1)}{2} = \frac{49(50)}{2}$.
6. Simplifying gives $x^2 - x + x^2 + x = 49 \times 50$, so $2x^2 = 2450$, which means $x^2 = 1225$.
7. Taking the positive square root, $x = 35$.

**Answer:** $35$

> Common mistake: Setting $S_{x-1} = S_x$ instead of equating the sum before $x$ to the sum after $x$.

### Question 5

*3 marks · Short answer*

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.

**Solution**

1. The volume of concrete for the first step is given by $\frac{1}{4} \times \frac{1}{2} \times 50 = \frac{25}{4}\text{ m}^3$.
2. The volume of concrete for the second step is $\frac{2}{4} \times \frac{1}{2} \times 50 = \frac{50}{4}\text{ m}^3$, and for the third step is $\frac{3}{4} \times \frac{1}{2} \times 50 = \frac{75}{4}\text{ m}^3$.
3. The volumes of concrete required for the 15 steps form an AP with first term $a = \frac{25}{4}$ and common difference $d = \frac{50}{4} - \frac{25}{4} = \frac{25}{4}$.
4. The total volume of concrete is the sum of the first 15 terms of this AP, $S_{15} = \frac{15}{2}[2a + (15 - 1)d]$.
5. Substituting the values, $S_{15} = \frac{15}{2}[2(\frac{25}{4}) + 14(\frac{25}{4})] = \frac{15}{2} \times \frac{25}{4} \times (2 + 14) = \frac{15}{2} \times \frac{25}{4} \times 16$.
6. Simplifying gives $S_{15} = 15 \times 25 \times 2 = 750\text{ m}^3$.

**Answer:** $750\text{ m}^3$

> Common mistake: Arithmetic errors when multiplying fractions in the sum formula.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions?

This chapter for the 2026-27 session contains four exercises in total. Exercise 5.1 has 4 questions, Exercise 5.2 has 20 questions, Exercise 5.3 has 20 questions, and the optional Exercise 5.4 has 5 questions. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page only.

### What topics do the questions cover in Class 10 Maths Chapter 5 Arithmetic Progressions?

The questions cover checking real-life situations for AP, finding the first term and common difference, and calculating the $nth$ term using $a_n = a + (n-1)d$. They also include finding the sum of $n$ terms using $S_n$ formulas and solving word problems related to geometry and rungs of a ladder. SwaVid's solutions on this page explain each of these concepts clearly.

### Which are the hardest question types in this chapter and how should I approach them?

Word problems involving $S_n$ and complex conditions on terms of two APs are often considered tricky by students. To approach them, first list the given values of $a$, $d$, $a_n$, or $S_n$ carefully and form linear equations. SwaVid's free PDF on this page breaks down these difficult problems into simple, manageable steps.

### How should I write my answers in the board exams to score full marks in Arithmetic Progressions?

You should always write down the given formulas clearly before substituting the values, such as writing the general term formula before solving for $n$. Showing proper working for linear equations in $a$ and $d$ ensures you do not lose marks for steps. Referring to SwaVid's detailed solutions on this page will help you learn the exact presentation format required.

### Is a free PDF for Class 10 Maths Chapter 5 Arithmetic Progressions available here?

Yes, you can access the complete and reliable NCERT solutions for this chapter in a free PDF format. SwaVid provides accurate step-by-step answers aligned with the textbook for the 2026-27 session on this page only to help with your exam preparation.

## Related pages

- [Exercise 5.1 solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-1)
- [Exercise 5.2 solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-2)
- [Exercise 5.3 solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-3)
- [Exercise 5.4 solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-4)
- [Arithmetic Progressions: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/arithmetic-progressions)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

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