Rbse Solutions for Class 10 Maths Chapter 5 Exercise 5.3 |Sum of n Terms of an AP

The formulas for the sum of the first $n$ terms of an AP are:

1. $$S_n = \frac{n}{2}[2a + (n-1)d]$$
2. $$S_n = \frac{n}{2}[a + l]$$where $a$ is the first term, $d$ is the common difference, $n$ is the number of terms, and $l$ (or $a_n$) is the last term.

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1. Find the Sum of the following APs

(i) $2, 7, 12, \dots$, to 10 terms.

$a=2, d=5, n=10$.

$$S_{10} = \frac{10}{2}[2(2) + (10-1)5]$$

$$S_{10} = 5[4 + 9(5)] = 5[4 + 45] = 5(49) = \mathbf{245}$$

(ii) $-37, -33, -29, \dots$, to 12 terms.

$a=-37, d = -33 – (-37) = 4, n=12$.

$$S_{12} = \frac{12}{2}[2(-37) + (12-1)4]$$

$$S_{12} = 6[-74 + 11(4)] = 6[-74 + 44] = 6(-30) = \mathbf{-180}$$

(iii) $0.6, 1.7, 2.8, \dots$, to 100 terms.

$a=0.6, d = 1.7 – 0.6 = 1.1, n=100$.

$$S_{100} = \frac{100}{2}[2(0.6) + (100-1)1.1]$$

$$S_{100} = 50[1.2 + 99(1.1)] = 50[1.2 + 108.9]$$

$$S_{100} = 50[110.1] = \mathbf{5505}$$

(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to 11 terms.

$a=\frac{1}{15}, n=11$.$d = \frac{1}{12} – \frac{1}{15} = \frac{5 – 4}{60} = \frac{1}{60}$.

$$S_{11} = \frac{11}{2}\left[2\left(\frac{1}{15}\right) + (11-1)\left(\frac{1}{60}\right)\right]$$

$$S_{11} = \frac{11}{2}\left[\frac{2}{15} + 10\left(\frac{1}{60}\right)\right] = \frac{11}{2}\left[\frac{2}{15} + \frac{1}{6}\right]$$

$$S_{11} = \frac{11}{2}\left[\frac{4 + 5}{30}\right] = \frac{11}{2}\left[\frac{9}{30}\right] = \frac{11}{2}\left[\frac{3}{10}\right] = \mathbf{\frac{33}{20}}$$

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2. Find the Sums given below

First, we find the number of terms ($n$) using $a_n = a + (n-1)d$. Then, we use $S_n = \frac{n}{2}[a + a_n]$.

(i) $7 + 10\frac{1}{2} + 14 + \dots + 84$

$a=7, a_n=84, d = 10.5 – 7 = 3.5$.

1. Find $n$:$$84 = 7 + (n-1)3.5$$$$77 = 3.5(n-1)$$$$n – 1 = \frac{77}{3.5} = 22 \implies n = 23$$
2. Find $S_{23}$:$$S_{23} = \frac{23}{2}[7 + 84] = \frac{23}{2}(91) = \frac{2093}{2} = \mathbf{1046\frac{1}{2}}$$

(ii) $34 + 32 + 30 + \dots + 10$

$a=34, a_n=10, d = 32 – 34 = -2$.

1. Find $n$:$$10 = 34 + (n-1)(-2)$$$$-24 = -2(n-1)$$$$12 = n – 1 \implies n = 13$$
2. Find $S_{13}$:$$S_{13} = \frac{13}{2}[34 + 10] = \frac{13}{2}(44) = 13(22) = \mathbf{286}$$

(iii) $-5 + (-8) + (-11) + \dots + (-230)$

$a=-5, a_n=-230, d = -8 – (-5) = -3$.

1. Find $n$:$$-230 = -5 + (n-1)(-3)$$$$-225 = -3(n-1)$$$$75 = n – 1 \implies n = 76$$
2. Find $S_{76}$:$$S_{76} = \frac{76}{2}[-5 + (-230)] = 38[-235] = \mathbf{-8930}$$

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3. In an AP (Solving for $a, d, n, a_n, S_n$)

(i) Given $a = 5, d = 3, a_n = 50$, find $n$ and $S_n$.

1. Find $n$: $50 = 5 + (n-1)3 \implies 45 = 3(n-1) \implies 15 = n – 1 \implies \mathbf{n = 16}$.
2. Find $S_{16}$: $S_{16} = \frac{16}{2}[5 + 50] = 8(55) = \mathbf{440}$.

(ii) Given $a = 7, a_{13} = 35$, find $d$ and $S_{13}$.

1. Find $d$: $a_{13} = a + 12d \implies 35 = 7 + 12d \implies 28 = 12d \implies \mathbf{d = \frac{28}{12} = \frac{7}{3}}$.
2. Find $S_{13}$: $S_{13} = \frac{13}{2}[7 + 35] = \frac{13}{2}(42) = 13(21) = \mathbf{273}$.

(iii) Given $a_{12} = 37, d = 3$, find $a$ and $S_{12}$.

1. Find $a$: $a_{12} = a + 11d \implies 37 = a + 11(3) \implies 37 = a + 33 \implies \mathbf{a = 4}$.
2. Find $S_{12}$: $S_{12} = \frac{12}{2}[4 + 37] = 6(41) = \mathbf{246}$.

(iv) Given $a_3 = 15, S_{10} = 125$, find $d$ and $a_{10}$.

1. Use $a_3$ and $S_{10}$ to find $a$ and $d$:
   • $a_3 \implies a + 2d = 15 \quad \dots (1)$
   • $S_{10} = \frac{10}{2}[2a + 9d] \implies 125 = 5(2a + 9d) \implies 25 = 2a + 9d \quad \dots (2)$
2. Solve the system: Multiply (1) by 2: $2a + 4d = 30 \quad \dots (3)$Subtract (3) from (2): $(2a + 9d) – (2a + 4d) = 25 – 30 \implies 5d = -5 \implies \mathbf{d = -1}$.
3. Find $a$: Substitute $d=-1$ into (1): $a + 2(-1) = 15 \implies a – 2 = 15 \implies a = 17$.
4. Find $a_{10}$: $a_{10} = a + 9d = 17 + 9(-1) = 17 – 9 = \mathbf{8}$.

(v) Given $d = 5, S_9 = 75$, find $a$ and $a_9$.

1. Find $a$: $S_9 = \frac{9}{2}[2a + (9-1)5] \implies 75 = \frac{9}{2}[2a + 40]$$$75 = 9a + 9(20) \implies 75 = 9a + 180$$$$9a = 75 – 180 = -105 \implies \mathbf{a = -\frac{105}{9} = -\frac{35}{3}}$$
2. Find $a_9$: $a_9 = a + 8d = -\frac{35}{3} + 8(5) = -\frac{35}{3} + 40 = \frac{-35 + 120}{3} = \mathbf{\frac{85}{3}}$.

(vi) Given $a = 2, d = 8, S_n = 90$, find $n$ and $a_n$.

1. Find $n$: $S_n = \frac{n}{2}[2(2) + (n-1)8] \implies 90 = \frac{n}{2}[4 + 8n – 8]$$$90 = \frac{n}{2}[8n – 4] = n(4n – 2)$$$$90 = 4n^2 – 2n \implies 4n^2 – 2n – 90 = 0$$Divide by 2: $2n^2 – n – 45 = 0$.Factor the quadratic: $(2n + 9)(n – 5) = 0$.Since $n$ must be a positive integer, $n \neq -9/2$. $\mathbf{n = 5}$.
2. Find $a_5$: $a_5 = a + 4d = 2 + 4(8) = 2 + 32 = \mathbf{34}$.

(vii) Given $a = 8, a_n = 62, S_n = 210$, find $n$ and $d$.

1. Find $n$: $S_n = \frac{n}{2}[a + a_n] \implies 210 = \frac{n}{2}[8 + 62]$$$210 = \frac{n}{2}(70) = 35n \implies n = \frac{210}{35} = \mathbf{6}$$
2. Find $d$: $a_6 = a + 5d \implies 62 = 8 + 5d$$$54 = 5d \implies \mathbf{d = \frac{54}{5} = 10.8}$$

(viii) Given $a_n = 4, d = 2, S_n = -14$, find $n$ and $a$.

1. Use $a_n$ and $S_n$:
   • $a_n = a + (n-1)2 \implies 4 = a + 2n – 2 \implies a = 6 – 2n \quad \dots (1)$
   • $S_n = \frac{n}{2}[a + a_n] \implies -14 = \frac{n}{2}[a + 4]$$$-28 = n(a + 4) \quad \dots (2)$$
2. Substitute (1) into (2):$$-28 = n((6 – 2n) + 4)$$$$-28 = n(10 – 2n) = 10n – 2n^2$$$$2n^2 – 10n – 28 = 0$$Divide by 2: $n^2 – 5n – 14 = 0$.Factor: $(n – 7)(n + 2) = 0$.Since $n$ must be positive, $\mathbf{n = 7}$.
3. Find $a$: Substitute $n=7$ into (1): $a = 6 – 2(7) = 6 – 14 \implies \mathbf{a = -8}$.

(ix) Given $a = 3, n = 8, S = 192$, find $d$.

1. Find $d$: $S_8 = \frac{8}{2}[2(3) + (8-1)d] \implies 192 = 4[6 + 7d]$$$\frac{192}{4} = 6 + 7d$$$$48 = 6 + 7d \implies 42 = 7d \implies \mathbf{d = 6}$$

(x) Given $l = 28, S = 144$, and $n = 9$, find $a$.

1. Find $a$: $S_9 = \frac{9}{2}[a + l]$$$144 = \frac{9}{2}[a + 28]$$$$144 \times \frac{2}{9} = a + 28$$$$16 \times 2 = a + 28 \implies 32 = a + 28 \implies \mathbf{a = 4}$$

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

$a=9, d=8, S_n=636$. We need $n$.

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$636 = \frac{n}{2}[2(9) + (n-1)8]$$

$$1272 = n[18 + 8n – 8]$$

$$1272 = n[8n + 10] = 8n^2 + 10n$$

$$8n^2 + 10n – 1272 = 0$$

Divide by 2: $4n^2 + 5n – 636 = 0$.

Using factorization (or quadratic formula): $4n^2 + 53n – 48n – 636 = 0$. (Split $5n$ into $53n – 48n$)

$$4n(n – 12) + 53(n – 12) = 0$$

$$(n – 12)(4n + 53) = 0$$

Since $n$ must be a positive integer, $n \neq -53/4$. $\mathbf{n = 12}$.

The number of terms is $\mathbf{12}$.

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5. First term $a=5$, last term $l=45$, sum $S_n=400$. Find $n$ and $d$.

1. Find $n$: $S_n = \frac{n}{2}[a + l] \implies 400 = \frac{n}{2}[5 + 45]$$$400 = \frac{n}{2}(50) = 25n \implies n = \frac{400}{25} = \mathbf{16}$$
2. Find $d$: $a_{16} = a + 15d \implies 45 = 5 + 15d$$$40 = 15d \implies d = \frac{40}{15} = \mathbf{\frac{8}{3}}$$

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6. First term $a=17$, last term $l=350$, common difference $d=9$. Find $n$ and $S_n$.

1. Find $n$: $l = a + (n-1)d \implies 350 = 17 + (n-1)9$$$333 = 9(n-1)$$$$n – 1 = \frac{333}{9} = 37 \implies n = 38$$
2. Find $S_{38}$: $S_{38} = \frac{38}{2}[17 + 350] = 19(367) = \mathbf{6973}$The number of terms is $\mathbf{38}$ and the sum is $\mathbf{6973}$.

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7. Sum of first 22 terms ($S_{22}$) where $d=7$ and $a_{22}=149$.

We can use $S_n = \frac{n}{2}[a + a_n]$. We first need $a$.

1. Find $a$: $a_{22} = a + 21d \implies 149 = a + 21(7)$$$149 = a + 147 \implies a = 2$$
2. Find $S_{22}$: $S_{22} = \frac{22}{2}[2 + 149] = 11(151) = \mathbf{1661}$

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8. Sum of first 51 terms ($S_{51}$) where $a_2=14$ and $a_3=18$.

1. Find $d$ and $a$:$$d = a_3 – a_2 = 18 – 14 = 4$$$$a = a_2 – d = 14 – 4 = 10$$
2. Find $S_{51}$:$$S_{51} = \frac{51}{2}[2a + 50d] = \frac{51}{2}[2(10) + 50(4)]$$$$S_{51} = \frac{51}{2}[20 + 200] = \frac{51}{2}(220) = 51(110) = \mathbf{5610}$$

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9. Sum of first $n$ terms ($S_n$)

Given $S_7 = 49$ and $S_{17} = 289$.

1. Use $S_n$ formula to find $a$ and $d$:
   • $S_7 = \frac{7}{2}[2a + 6d] = 49 \implies 7(a + 3d) = 49 \implies a + 3d = 7 \quad \dots (1)$
   • $S_{17} = \frac{17}{2}[2a + 16d] = 289 \implies 17(a + 8d) = 289 \implies a + 8d = 17 \quad \dots (2)$
2. Solve the system: Subtract (1) from (2):$$(a + 8d) – (a + 3d) = 17 – 7 \implies 5d = 10 \implies d = 2$$Substitute $d=2$ into (1): $a + 3(2) = 7 \implies a = 1$.
3. Find $S_n$:$$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(1) + (n-1)2]$$$$S_n = \frac{n}{2}[2 + 2n – 2] = \frac{n}{2}[2n] = \mathbf{n^2}$$

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10. Show $a_n$ forms an AP and find $S_{15}$

An AP is formed if the difference between $a_{n+1}$ and $a_n$ is constant ($d$).

(i) $a_n = 3 + 4n$

1. Show AP:$$a_{n+1} – a_n = [3 + 4(n+1)] – [3 + 4n]$$$$= (3 + 4n + 4) – (3 + 4n) = 4$$Since the difference is constant, $\mathbf{d=4}$. The sequence forms an AP.
2. Find $a$ and $S_{15}$:
   • $a = a_1 = 3 + 4(1) = 7$
   • $a_{15} = 3 + 4(15) = 63$ (or use $a_{15} = a + 14d = 7 + 14(4) = 63$)
   • $S_{15} = \frac{15}{2}[a + a_{15}] = \frac{15}{2}[7 + 63] = \frac{15}{2}(70) = 15(35) = \mathbf{525}$

(ii) $a_n = 9 – 5n$

1. Show AP:$$a_{n+1} – a_n = [9 – 5(n+1)] – [9 – 5n]$$$$= (9 – 5n – 5) – (9 – 5n) = -5$$Since the difference is constant, $\mathbf{d=-5}$. The sequence forms an AP.
2. Find $a$ and $S_{15}$:
   • $a = a_1 = 9 – 5(1) = 4$
   • $a_{15} = 9 – 5(15) = 9 – 75 = -66$
   • $S_{15} = \frac{15}{2}[4 + (-66)] = \frac{15}{2}(-62) = 15(-31) = \mathbf{-465}$

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11. Given $S_n = 4n – n^2$

• First term ($S_1$): $S_1 = 4(1) – (1)^2 = 4 – 1 = \mathbf{3}$.
• Sum of first two terms ($S_2$): $S_2 = 4(2) – (2)^2 = 8 – 4 = \mathbf{4}$.
• Second term ($a_2$): $a_2 = S_2 – S_1 = 4 – 3 = \mathbf{1}$.
• Third term ($a_3$): First find $S_3$: $S_3 = 4(3) – (3)^2 = 12 – 9 = 3$.$$a_3 = S_3 – S_2 = 3 – 4 = \mathbf{-1}$$
• Tenth term ($a_{10}$): $a_{10} = S_{10} – S_9$.
  • $S_{10} = 4(10) – (10)^2 = 40 – 100 = -60$.
  • $S_9 = 4(9) – (9)^2 = 36 – 81 = -45$.$$a_{10} = -60 – (-45) = \mathbf{-15}$$
• $n^{th}$ term ($a_n$): $a_n = S_n – S_{n-1}$.$$S_{n-1} = 4(n-1) – (n-1)^2 = 4n – 4 – (n^2 – 2n + 1) = -n^2 + 6n – 5$$$$a_n = (4n – n^2) – (-n^2 + 6n – 5)$$$$a_n = 4n – n^2 + n^2 – 6n + 5 = \mathbf{5 – 2n}$$

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12. Sum of the first 40 positive integers divisible by 6.

This AP is $6, 12, 18, \dots$. $a=6, d=6, n=40$.

$$S_{40} = \frac{40}{2}[2(6) + (40-1)6]$$

$$S_{40} = 20[12 + 39(6)] = 20[12 + 234]$$

$$S_{40} = 20[246] = \mathbf{4920}$$

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13. Sum of the first 15 multiples of 8.

This AP is $8, 16, 24, \dots$. $a=8, d=8, n=15$.

$$S_{15} = \frac{15}{2}[2(8) + (15-1)8]$$

$$S_{15} = \frac{15}{2}[16 + 14(8)] = \frac{15}{2}[16 + 112]$$

$$S_{15} = \frac{15}{2}(128) = 15(64) = \mathbf{960}$$

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14. Sum of the odd numbers between 0 and 50.

The odd numbers are $1, 3, 5, \dots, 49$. $a=1, d=2, a_n=49$.

1. Find $n$: $49 = 1 + (n-1)2 \implies 48 = 2(n-1) \implies 24 = n – 1 \implies n = 25$.
2. Find $S_{25}$: $S_{25} = \frac{25}{2}[1 + 49] = \frac{25}{2}(50) = 25(25) = \mathbf{625}$.

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15. Penalty for delay

The penalties form an AP: $200, 250, 300, \dots$$a=200, d=50$. The contractor delayed by $n=30$ days. We need $S_{30}$.

$$S_{30} = \frac{30}{2}[2(200) + (30-1)50]$$

$$S_{30} = 15[400 + 29(50)] = 15[400 + 1450]$$

$$S_{30} = 15[1850] = \mathbf{27750}$$

The contractor has to pay $\mathbf{₹ 27,750}$ as penalty.

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16. Cash Prizes

Let the first prize be $a$. The prizes form an AP: $a, a-20, a-40, \dots$$d=-20$. Total prizes $n=7$. Total sum $S_7=700$. We need $a$ and the list of prizes.

$$S_7 = \frac{7}{2}[2a + (7-1)(-20)]$$

$$700 = \frac{7}{2}[2a – 120]$$

$$100 = \frac{1}{2}[2a – 120] = a – 60$$

$$a = 160$$

The value of the prizes are: $160, 140, 120, 100, 80, 60, 40$.

The values of the prizes are $\mathbf{₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, \text{ and } ₹ 40}$.

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17. Planting Trees

The number of trees planted by each class is:

• Class I: 1 tree $\times$ 3 sections = 3 trees
• Class II: 2 trees $\times$ 3 sections = 6 trees
• Class III: 3 trees $\times$ 3 sections = 9 trees$\dots$
• Class XII: 12 trees $\times$ 3 sections = 36 treesThe total number of trees forms an AP: $3, 6, 9, \dots, 36$.$a=3, d=3, n=12$ (for classes I to XII).$$S_{12} = \frac{12}{2}[a + l] = 6[3 + 36] = 6(39) = \mathbf{234}$$The students will plant $\mathbf{234}$ trees.

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18. Length of a Spiral

The lengths of the successive semicircles are $l_1, l_2, l_3, \dots$

The length of a semicircle is $\pi r$.

Radii $r$: $0.5, 1.0, 1.5, 2.0, \dots$

Lengths $l$: $\pi(0.5), \pi(1.0), \pi(1.5), \pi(2.0), \dots$

This is an AP with $a = 0.5\pi, d = 0.5\pi$. $n=13$.

$$S_{13} = \frac{13}{2}[2a + (13-1)d]$$

$$S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)]$$

$$S_{13} = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi)$$

Substitute $\pi = 22/7$:

$$S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7} = 13 \times \frac{22}{2} = 13 \times 11 = \mathbf{143}$$

The total length of the spiral is $\mathbf{143 \text{ cm}}$.

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19. Stacking Logs

The logs form an AP: $20, 19, 18, \dots$.$a=20, d=-1$. Total number of logs $S_n=200$. We need $n$ (number of rows) and $a_n$ (logs in the top row).

1. Find $n$:$$S_n = \frac{n}{2}[2(20) + (n-1)(-1)]$$$$200 = \frac{n}{2}[40 – n + 1] \implies 400 = n[41 – n]$$$$400 = 41n – n^2$$$$n^2 – 41n + 400 = 0$$Factor: $(n – 16)(n – 25) = 0$.Possible values for $n$ are $16$ and $25$.
2. Check $a_n$ for both values:
   • If $n=25$: $a_{25} = 20 + (25-1)(-1) = 20 – 24 = -4$. This is not possible (number of logs cannot be negative).
   • If $n=16$: $a_{16} = 20 + (16-1)(-1) = 20 – 15 = 5$. This is possible.The number of rows is $\mathbf{16}$ and the number of logs in the top row is $\mathbf{5}$.

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20. Potato Race

The distances run for each potato form a sequence. The distance to pick up and drop the $n^{th}$ potato is $2 \times (\text{distance to } n^{th} \text{ potato})$.

• Distance to 1st potato: 5 m. Distance run: $2 \times 5 = 10$ m.
• Distance to 2nd potato: $5 + 3 = 8$ m. Distance run: $2 \times 8 = 16$ m.
• Distance to 3rd potato: $5 + 3 + 3 = 11$ m. Distance run: $2 \times 11 = 22$ m.The distances run for each potato form an AP: $10, 16, 22, \dots$$a=10, d=6$. Total number of potatoes $n=10$. We need $S_{10}$.$$S_{10} = \frac{10}{2}[2(10) + (10-1)6]$$$$S_{10} = 5[20 + 9(6)] = 5[20 + 54]$$$$S_{10} = 5[74] = \mathbf{370}$$The total distance the competitor has to run is $\mathbf{370 \text{ meters}}$.

Last Updated on November 27, 2025 by Aman Singh