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

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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.
Rbse Solutions for Class 10 Maths Exercise 5.1
Rbse Solutions for Class 10 Maths Exercise 5.2
Rbse Solutions for Class 10 Maths Exercise 5.3
Rbse Solutions for Class 10 Maths Chapter 5 Exercise 5.3 |Sum of n Terms of an AP
Rbse Solutions for Class 10 Maths Chapter 5 Exercise 5.3 |Sum of n Terms of an AP
Rbse Solutions for Class 10 Maths Chapter 5 Exercise 5.3 |Sum of n Terms of an AP
Rbse Solutions for Class 10 Maths Chapter 5 Exercise 5.3 |Sum of n Terms of an AP

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.

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

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

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

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

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 $S_{76}$:$$S_{76} = \frac{76}{2}[-5 + (-230)] = 38[-235] = \mathbf{-8930}$$

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]$$$144 = \frac{9}{2}[a + 28]$$$$144 \times \frac{2}{9} = a + 28$$$$16 \times 2 = a + 28 \implies 32 = a + 28 \implies \mathbf{a = 4}$$

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 $\mathbf{12}$.

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$$$40 = 15d \implies d = \frac{40}{15} = \mathbf{\frac{8}{3}}$$

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 $\mathbf{6973}$.

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 $S_{22}$: $S_{22} = \frac{22}{2}[2 + 149] = 11(151) = \mathbf{1661}$

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 $S_{51}$:$$S_{51} = \frac{51}{2}[2a + 50d] = \frac{51}{2}[2(10) + 50(4)]$$$$S_{51} = \frac{51}{2}[20 + 200] = \frac{51}{2}(220) = 51(110) = \mathbf{5610}$$

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

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 $S_{15}$:
    • $a = a_1 = 9 – 5(1) = 4$
    • $a_{15} = 9 – 5(15) = 9 – 75 = -66$
    • $S_{15} = \frac{15}{2}[4 + (-66)] = \frac{15}{2}(-62) = 15(-31) = \mathbf{-465}$

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 ($a_n$): $a_n = S_n – S_{n-1}$.$$S_{n-1} = 4(n-1) – (n-1)^2 = 4n – 4 – (n^2 – 2n + 1) = -n^2 + 6n – 5$$$$a_n = (4n – n^2) – (-n^2 + 6n – 5)$$$$a_n = 4n – n^2 + n^2 – 6n + 5 = \mathbf{5 – 2n}$$

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

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

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

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

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

13. Sum of the first 15 multiples of 8.

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

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

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

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

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 $S_{25}$: $S_{25} = \frac{25}{2}[1 + 49] = \frac{25}{2}(50) = 25(25) = \mathbf{625}$.

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.

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 $\mathbf{₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, \text{ and } ₹ 40}$.

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.

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 $\mathbf{143 \text{ cm}}$.

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 $\mathbf{5}$.

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 $\mathbf{370 \text{ meters}}$.
Rbse Solutions for Class 10 Maths Exercise 5.1
Rbse Solutions for Class 10 Maths Exercise 5.2
Rbse Solutions for Class 10 Maths Exercise 5.3

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