---
title: "NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3"
url: https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-3
dateModified: 2026-10-07T15:51:43+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3

Chapter 5: Arithmetic Progressions. Every question from Exercise 5.3, 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 5.3

### 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.

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