NCERT Class 10 Maths – Exercise 5.3 Solutions

NCERT Class 10 Maths

Chapter 5 – Arithmetic Progressions | Exercise 5.3

(Rationalized Syllabus 2025-26)

💡 Key Formulas for Sum of AP

The sum of first $n$ terms $S_n$ is given by:

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

Or, if the last term $l$ (or $a_n$) is known:

$$S_n = \frac{n}{2}(a + l)$$

Find the sum of the following APs:

(i) 2, 7, 12, … to 10 terms.

$a=2, d=7-2=5, n=10$.

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

$$= 5[4 + 9(5)] = 5[4 + 45] = 5 \times 49 = 245$$

✔️ 245

(ii) -37, -33, -29, … to 12 terms.

$a=-37, d=-33 – (-37) = 4, n=12$.

$$S_{12} = \frac{12}{2}[2(-37) + (11)(4)]$$

$$= 6[-74 + 44] = 6(-30) = -180$$

✔️ -180

(iii) 0.6, 1.7, 2.8, … to 100 terms.

$a=0.6, d=1.1, n=100$.

$$S_{100} = \frac{100}{2}[2(0.6) + (99)(1.1)]$$

$$= 50[1.2 + 108.9] = 50[110.1] = 5505$$

✔️ 5505

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

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

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

$$= \frac{11}{2}[\frac{2}{15} + \frac{1}{6}] = \frac{11}{2}[\frac{4+5}{30}] = \frac{11}{2} \times \frac{9}{30} = \frac{33}{20}$$

✔️ 33/20

Find the sums given below:

(i) $7 + 10\frac{1}{2} + 14 + … + 84$

Here $a=7, d=3.5 (\frac{7}{2}), l=84$. First find $n$.

$$84 = 7 + (n-1)\frac{7}{2} \Rightarrow 77 = (n-1)\frac{7}{2} \Rightarrow n-1 = 22 \Rightarrow n=23$$

$$S_{23} = \frac{23}{2}(7 + 84) = \frac{23}{2}(91) = \frac{2093}{2} = 1046\frac{1}{2}$$

✔️ 1046 1/2

(ii) $34 + 32 + 30 + … + 10$

$a=34, d=-2, l=10$.

$$10 = 34 + (n-1)(-2) \Rightarrow -24 = -2(n-1) \Rightarrow n=13$$

$$S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2}(44) = 13 \times 22 = 286$$

✔️ 286

(iii) $-5 + (-8) + (-11) + … + (-230)$

$a=-5, d=-3, l=-230$.

$$-230 = -5 + (n-1)(-3) \Rightarrow -225 = -3(n-1) \Rightarrow n=76$$

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

✔️ -8930

In an AP:

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

$50 = 5 + (n-1)3 \Rightarrow 45 = 3(n-1) \Rightarrow n = 16$.

$S_{16} = \frac{16}{2}(5+50) = 8(55) = 440$.

✔️ n = 16, Sₙ = 440

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

$35 = 7 + 12d \Rightarrow 28 = 12d \Rightarrow d = 28/12 = 7/3$.

$S_{13} = \frac{13}{2}(7+35) = \frac{13}{2}(42) = 13 \times 21 = 273$.

✔️ d = 7/3, S₁₃ = 273

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

$37 = a + 11(3) \Rightarrow a = 37 – 33 = 4$.

$S_{12} = \frac{12}{2}(4+37) = 6(41) = 246$.

✔️ a = 4, S₁₂ = 246

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

Eq 1: $a + 2d = 15$.

Eq 2: $125 = \frac{10}{2}[2a + 9d] \Rightarrow 125 = 5(2a+9d) \Rightarrow 2a + 9d = 25$.

Solving system: Multiply Eq 1 by 2 $\Rightarrow 2a+4d=30$. Subtract from Eq 2 $\Rightarrow 5d = -5 \Rightarrow d = -1$.

$a = 15 – 2(-1) = 17$.

$a_{10} = 17 + 9(-1) = 8$.

✔️ d = -1, a₁₀ = 8

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

$75 = \frac{9}{2}[2a + 8(5)] \Rightarrow \frac{150}{9} = 2a + 40 \Rightarrow \frac{50}{3} = 2a + 40$.

$2a = \frac{50}{3} – 40 = \frac{50-120}{3} = \frac{-70}{3} \Rightarrow a = \frac{-35}{3}$.

$a_9 = \frac{-35}{3} + 8(5) = \frac{-35+120}{3} = \frac{85}{3}$.

✔️ a = -35/3, a₉ = 85/3

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

$90 = \frac{n}{2}[4 + (n-1)8] \Rightarrow 180 = n(8n – 4) \Rightarrow 180 = 4n(2n-1) \Rightarrow 45 = 2n^2 – n$.

Quadratic: $2n^2 – n – 45 = 0$. Factors of -90 summing to -1: -10, 9.

$2n^2 – 10n + 9n – 45 = 0 \Rightarrow 2n(n-5) + 9(n-5) = 0$.

$n=5$ (since n cannot be negative/fraction).

$a_5 = 2 + 4(8) = 34$.

✔️ n = 5, aₙ = 34

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

$S_n = \frac{n}{2}(a + a_n) \Rightarrow 210 = \frac{n}{2}(8 + 62) \Rightarrow 420 = 70n \Rightarrow n = 6$.

$62 = 8 + 5d \Rightarrow 54 = 5d \Rightarrow d = 54/5$.

✔️ n = 6, d = 54/5

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

$a_n = a + (n-1)2 = 4 \Rightarrow a + 2n – 2 = 4 \Rightarrow a = 6 – 2n$.

$S_n = \frac{n}{2}(a+4) = -14 \Rightarrow n(6-2n+4) = -28$.

$n(10-2n) = -28 \Rightarrow 10n – 2n^2 = -28 \Rightarrow 2n^2 – 10n – 28 = 0$.

$n^2 – 5n – 14 = 0 \Rightarrow (n-7)(n+2) = 0$. So $n=7$.

$a = 6 – 2(7) = -8$.

✔️ n = 7, a = -8

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

$192 = \frac{8}{2}[2(3) + 7d] \Rightarrow 192 = 4[6 + 7d]$.

$48 = 6 + 7d \Rightarrow 42 = 7d \Rightarrow d = 6$.

✔️ d = 6

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

$144 = \frac{9}{2}(a + 28) \Rightarrow 288 = 9(a + 28) \Rightarrow 32 = a + 28 \Rightarrow a = 4$.

✔️ a = 4

Q4-Q10

Application Problems

Q4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?

$a=9, d=8, S_n=636$.

$$636 = \frac{n}{2}[18 + (n-1)8] \Rightarrow 1272 = n(18 + 8n – 8)$$

$$1272 = n(10 + 8n) \Rightarrow 1272 = 10n + 8n^2$$

$$8n^2 + 10n – 1272 = 0 \Rightarrow 4n^2 + 5n – 636 = 0$$

Solving quadratic: $n = \frac{-5 \pm \sqrt{25 – 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8}$.

Taking positive: $n = 96/8 = 12$.

✔️ 12 terms

Q5. First term is 5, last term is 45, sum is 400. Find $n$ and $d$.

$400 = \frac{n}{2}(5 + 45) \Rightarrow 800 = 50n \Rightarrow n = 16$.

$45 = 5 + 15d \Rightarrow 40 = 15d \Rightarrow d = 40/15 = 8/3$.

✔️ n = 16, d = 8/3

Q6. First and last terms are 17 and 350. $d=9$. Find $n$ and sum.

$350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow n-1 = 37 \Rightarrow n = 38$.

$S_{38} = \frac{38}{2}(17 + 350) = 19(367) = 6973$.

✔️ n = 38, Sum = 6973

Q7. Find sum of first 22 terms where $d=7$ and 22nd term is 149.

$149 = a + 21(7) \Rightarrow 149 = a + 147 \Rightarrow a = 2$.

$S_{22} = \frac{22}{2}(2 + 149) = 11(151) = 1661$.

✔️ 1661

Q8. Find sum of first 51 terms where 2nd term is 14 and 3rd term is 18.

$d = 18 – 14 = 4$.

$a = 14 – 4 = 10$.

$S_{51} = \frac{51}{2}[20 + 50(4)] = \frac{51}{2}(220) = 51(110) = 5610$.

✔️ 5610

Q9. If sum of 7 terms is 49 and 17 terms is 289, find sum of $n$ terms.

Notice pattern: $S_7 = 7^2$, $S_{17} = 17^2$.

This is the property of sum of first $n$ odd numbers.

$S_n = n^2$.

✔️ n²

Q10. Show that $a_n = 3 + 4n$ forms an AP and find $S_{15}$.

$a_1 = 7, a_2 = 11, a_3 = 15$. $d=4$. It is an AP.

$S_{15} = \frac{15}{2}[2(7) + 14(4)] = \frac{15}{2}(14 + 56) = \frac{15}{2}(70) = 15(35) = 525$.

✔️ 525

Q11-Q15

Logic Problems

Q11. If $S_n = 4n – n^2$, find $a_1, a_2, a_3, a_{10}, a_n$.

$S_1 = 4(1) – 1 = 3 \Rightarrow a_1 = 3$.

$S_2 = 4(2) – 4 = 4 \Rightarrow a_2 = S_2 – S_1 = 4 – 3 = 1$.

$S_3 = 12 – 9 = 3 \Rightarrow a_3 = S_3 – S_2 = 3 – 4 = -1$.

AP is $3, 1, -1$. ($d = -2$).

$a_{10} = 3 + 9(-2) = 3 – 18 = -15$.

$a_n = 3 + (n-1)(-2) = 3 – 2n + 2 = 5 – 2n$.

✔️ a₁=3, a₂=1, a₃=-1, a₁₀=-15, aₙ=5-2n

Q12. Sum of first 40 positive integers divisible by 6.

$6, 12, 18, …$. $a=6, d=6, n=40$.

$S_{40} = \frac{40}{2}[12 + 39(6)] = 20[12 + 234] = 20(246) = 4920$.

✔️ 4920

Q13. Sum of first 15 multiples of 8.

$a=8, d=8, n=15$.

$S_{15} = \frac{15}{2}[16 + 14(8)] = \frac{15}{2}(16 + 112) = \frac{15}{2}(128) = 15(64) = 960$.

✔️ 960

Q14. Sum of odd numbers between 0 and 50.

$1, 3, 5, …, 49$.

$a=1, l=49$. Number of terms $n=25$.

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

✔️ 625

Q15. Penalty for delay: ₹200, ₹250, ₹300… for 30 days. Find total penalty.

$a=200, d=50, n=30$.

$S_{30} = \frac{30}{2}[400 + 29(50)] = 15[400 + 1450] = 15(1850) = 27750$.

✔️ ₹27,750

Q16-Q20

Real Life Problems

Q16. Sum of ₹700 to be used for 7 cash prizes. Each prize is ₹20 less than preceding one. Find value of each prize.

$n=7, S_7=700, d=-20$.

$700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a – 120 \Rightarrow 2a = 320 \Rightarrow a = 160$.

Prizes: 160, 140, 120, 100, 80, 60, 40.

✔️ ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

Q17. Planting trees: Class I plants 1, Class II plants 2… till Class XII. There are 3 sections of each class. Total trees?

Trees per class group: $3 \times 1, 3 \times 2, …, 3 \times 12$.

Series: 3, 6, …, 36. $n=12$.

$S_{12} = \frac{12}{2}(3 + 36) = 6(39) = 234$.

✔️ 234 trees

Q18. Spiral made of semicircles with radii 0.5, 1.0, 1.5… Total length of 13 spirals?

Perimeter of semicircle = $\pi r$.

$l_1 = \pi(0.5)$, $l_2 = \pi(1)$. $d = 0.5\pi$. $n=13$.

Total = $S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi)$.

Using $\pi = 22/7$: Total = $\frac{13}{2} \times 7 \times \frac{22}{7} = 13 \times 11 = 143$.

✔️ 143 cm

Q19. 200 logs stacked: 20 bottom, 19 next, etc. How many rows and how many logs in top row?

$S_n=200, a=20, d=-1$.

$200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = n(41 – n) \Rightarrow n^2 – 41n + 400 = 0$.

$(n-16)(n-25) = 0$. So $n=16$ or $n=25$.

If $n=25$, $a_{25} = 20 – 24 = -4$ (Impossible).

So $n=16$. Top row ($a_{16}$) = $20 – 15 = 5$.

✔️ 16 rows, 5 logs in top row

Q20. Potato race: Bucket at 5m. Potatoes at 3m intervals. 10 potatoes. Total distance run?

Distance for 1st potato = $2 \times 5 = 10$.

Distance for 2nd potato = $2 \times (5+3) = 16$.

Series: 10, 16, 22… ($a=10, d=6, n=10$).

$S_{10} = \frac{10}{2}[20 + 9(6)] = 5[20 + 54] = 5(74) = 370$.

✔️ 370 m

🎉 Exercise 5.3 Completed | Chapter 5 Finished!