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NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3 are provided to help students strengthen their understanding of Arithmetic Progressions (AP) and prepare effectively for the Class 10 board exams. These solutions are prepared by subject experts and follow the latest CBSE syllabus and exam pattern.

Exercise 5.3 mainly focuses on finding the nth term of an Arithmetic Progression using the formula:

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

Arithmetic Progression

Medium

Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP:

Arithmetic Progression

Medium

\begin{array}{l} 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 . \end{array}

Arithmetic Progression

Difficult

In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.

[CBSE - 2024]

Arithmetic Progression

Medium

The sum of first and eighth terms of an A.P. is 32 and their product is 60 . Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

[CBSE - 2024]

Arithmetic Progression

Medium

If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

[CBSE - 2019]

Arithmetic Progression

Medium

\mathrm{S}_n,

the sum of first n terms of an AP is given by

\mathrm{S}_n=3 n^2-4 n,

find the

n^{\rm th}

term.

[CBSE - 2019]

Arithmetic Progression

Medium

Which term of the AP 3, 15, 27, 39, … will be 120 more than its 21st term?

[CBSE - 2019]

Arithmetic Progression

Easy

In an A.P., if the first term a = 7,

n^{th}

term

a_n=84

and the sum of first n terms

s_n=\frac{2093}{2}

, then n is equal to:

[CBSE - 2024]

Arithmetic Progression

Medium

If the sum of first m terms of an AP is

\mathrm{AP}\text{ is }2\mathrm{~m}^2+3\mathrm{~m}{,}

then its second term is:

[CBSE - 2025]

Arithmetic Progression

Medium

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 the following figure)

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?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Arithmetic Progression

Medium

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

Arithmetic Progression

Medium

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.

Arithmetic Progression

Medium

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

Arithmetic Progression

Medium

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

Arithmetic Progression

Medium

\begin{array}{l}{Choose the correct choice in the following and justify:}\\ {(i) 30th term of the AP: 10, 7, 4, . . . , is}\\ {     (A) 97     (B) 77      (C) }-{77      (D) }-{87}\\ {(ii) 11th term of the AP: }-{3, }\frac{-{1}}{2}{, 2, . . ., is}\\ {      (A) 28     (B) 22     (C) }-{38      (D) }-{48}\frac{1}{2}\end{array}

Arithmetic Progression

Medium

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

Arithmetic Progression

Medium

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Arithmetic Progression

Medium

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?

Arithmetic Progression

Medium

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.

Arithmetic Progression

Medium

Find the 20th term from the last term of the AP: 3, 8, 13, . . ., 253.

Arithmetic Progression

Medium

For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Arithmetic Progression

Medium

How many multiples of 4 lie between 10 and 250?

Arithmetic Progression

Medium

How many three-digit numbers are divisible by 7?

Arithmetic Progression

Medium

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Arithmetic Progression

Medium

If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?

Arithmetic Progression

Medium

\begin{array}{l} Find the number of terms in each of the following APs: \\ (i) 7, 13, 19, . . . , 205         (ii) 18, 15 \frac{1}{2}, 13, . . . , - 47 \end{array}

Arithmetic Progression

Medium

The sum of the first 'm' terms of an Arithmetic Progression (AP) is given by the expression

2m^2 + 3m

. Determine the second term of this AP.

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

$a_n = a + (n - 1)d$

$a_{n} = a + (n - 1) d$

In this exercise, students learn how to:

Find any particular term of an AP
Determine whether a given number is a term of an AP
Solve real-life problems using the concept of nth term

The solutions are explained in a clear, step-by-step manner, helping students understand the logic behind each problem.

Q1. Find the sum of the following APs.

(i) 2, 7, 12, … to 10 terms
a = 2, d = 5, n = 10

$S_{10}=\frac{10}{2}[2a+(10-1)d]=5[4+45]=245$

$S_{10} = 210 [2 a + (10 - 1) d] = 5 [4 + 45] = 245$

✅ Sum = 245

(ii) −37, −33, −29, … to 12 terms
a = −37, d = 4, n = 12

$S_{12}=\frac{12}{2}[2(-37)+(11)(4)]=6[-74+44]=6(-30)=-180$

$S_{12} = 212 [2 (- 37) + (11) (4)] = 6 [- 74 + 44] = 6 (- 30) = - 180$

✅ Sum = −180

(iii) 0.6, 1.7, 2.8, … to 100 terms
a = 0.6, d = 1.1, n = 100

$S_{100}=\frac{100}{2}[2(0.6)+99(1.1)]=50[1.2+108.9]=50(110.1)=5505$

$S_{100} = 2100 [2 (0.6) + 99 (1.1)] = 50 [1.2 + 108.9] = 50 (110.1) = 5505$

✅ Sum = 5505

(iv) 1/15, 1/12, 1/10, … to 11 terms
a = 1/15, d = 1/12 − 1/15 = 1/60, n = 11

$S_{11}=\frac{11}{2}\left[2\cdot\frac{1}{15}+10\cdot\frac{1}{60}\right] =\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right] =\frac{11}{2}\left[\frac{3}{10}\right]=\frac{33}{20}$

$S_{11} = 211 [2 \cdot 151 + 10 \cdot 601] = 211 [152 + 61] = 211 [103] = 2033$

✅ Sum = 33/20

Q2. Find the sums given below

(i) 7, 7

$\frac{1}{2}$

$2 1$ , 8, … , 84
a = 7, d = 1/2, l = 84

$84=7+(n-1)\frac{1}{2}\Rightarrow 77=\frac{n-1}{2}\Rightarrow n=155$

$84 = 7 + (n - 1) 21 \Rightarrow 77 = 2 n - 1 \Rightarrow n = 155$

$S=\frac{n}{2}(a+l)=\frac{155}{2}(7+84)=\frac{155}{2}\cdot 91=\frac{14105}{2}$

$S = 2 n (a + l) = 2155 (7 + 84) = 2155 \cdot 91 = 214105$

✅ Sum = 14105/2

(ii) 34 + 32 + 30 + … + 10
a = 34, d = −2, l = 10

$10=34+(n-1)(-2)\Rightarrow 10=34-2n+2\Rightarrow 2n=26\Rightarrow n=13$

$10 = 34 + (n - 1) (- 2) \Rightarrow 10 = 34 - 2 n + 2 \Rightarrow 2 n = 26 \Rightarrow n = 13$

$S=\frac{13}{2}(34+10)=\frac{13}{2}\cdot 44=286$

$S = 213 (34 + 10) = 213 \cdot 44 = 286$

✅ Sum = 286

(iii) −5 + (−8) + (−11) + … + (−230)
a = −5, d = −3, l = −230

$-230=-5+(n-1)(-3)\Rightarrow -225=-3(n-1)\Rightarrow n-1=75\Rightarrow n=76$

$- 230 = - 5 + (n - 1) (- 3) \Rightarrow - 225 = - 3 (n - 1) \Rightarrow n - 1 = 75 \Rightarrow n = 76$

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

$S = 276 (- 5 - 230) = 38 (- 235) = - 8930$

✅ Sum = −8930

Q3. In an AP

(i) a = 5, d = 3,

$a_n$

$a_{n}$ =50; find n and

$S_n$

$S_{n}$

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

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

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

$S_{16} = 216 (5 + 50) = 8 \cdot 55 = 440$

✅ n = 16,

$S_{16}=440$

$S_{16} = 440$

(ii)

$a_{13}=35$

$a_{13} = 35$ , a = 7; find d and

$S_{13}$

$35=7+12d\Rightarrow d=\frac{28}{12}=\frac{7}{3}$

$35 = 7 + 12 d \Rightarrow d = 1228 = 37$

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

$S_{13} = 213 (7 + 35) = 213 \cdot 42 = 273$

✅ d = 7/3,

$S_{13}=273$

$S_{13} = 273$

(iii)

$a_{12}=37$

$a_{12} = 37$ , d = 3; find a and

$S_{12}$

$37=a+11\cdot 3 \Rightarrow a=4$

$37 = a + 11 \cdot 3 \Rightarrow a = 4$

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

$S_{12} = 212 (4 + 37) = 6 \cdot 41 = 246$

✅ a = 4,

$S_{12}=246$

$S_{12} = 246$

(iv)

$a_3=15$

$a_{3} = 15$ ,

$S_{10}=125$

$S_{10} = 125$ ; find d and

$a_{10}$

$a+2d=15 \quad ...(1)$

$a + 2 d = 15 ... (1)$

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

$125 = 210 [2 a + 9 d] \Rightarrow 25 = 2 a + 9 d ... (2)$

From (1): 2a+4d=30
Subtract from (2): (2a+9d)−(2a+4d)=25−30 ⇒ 5d=−5 ⇒ d=−1
Then a=15−2(−1)=17

$a_{10}=a+9d=17-9=8$

$a_{10} = a + 9 d = 17 - 9 = 8$

✅ d = −1,

$a_{10}=8$

$a_{10} = 8$

(v) d = 5,

$S_9=75$

$S_{9} = 75$ ; find a and

$a_9$

$a_{9}$

$75=\frac{9}{2}[2a+8\cdot 5]\Rightarrow 75=\frac{9}{2}(2a+40) \Rightarrow 150=9(2a+40)\Rightarrow 2a+40=\frac{50}{3} \Rightarrow a=-\frac{35}{3}$

$75 = 29 [2 a + 8 \cdot 5] \Rightarrow 75 = 29 (2 a + 40) \Rightarrow 150 = 9 (2 a + 40) \Rightarrow 2 a + 40 = 350 \Rightarrow a = - 335$

$a_9=a+8d=-\frac{35}{3}+40=\frac{85}{3}$

$a_{9} = a + 8 d = - 335 + 40 = 385$

✅ a = −35/3,

$a_9=85/3$

$a_{9} = 85/3$

(vi) a = 2, d = 8,

$S_n=90$

$S_{n} = 90$ ; find n and

$a_n$

$a_{n}$

$90=\frac{n}{2}[2\cdot2+(n-1)8] =\frac{n}{2}[4+8n-8] =\frac{n}{2}(8n-4)=n(4n-2)$

$90 = 2 n [2 \cdot 2 + (n - 1) 8] = 2 n [4 + 8 n - 8] = 2 n (8 n - 4) = n (4 n - 2)$

$4n^2-2n-90=0 \Rightarrow 2n^2-n-45=0 \Rightarrow (2n+9)(n-5)=0 \Rightarrow n=5$

$4 n^{2} - 2 n - 90 = 0 \Rightarrow 2 n^{2} - n - 45 = 0 \Rightarrow (2 n + 9) (n - 5) = 0 \Rightarrow n = 5$

$a_5=2+4\cdot 8=34$

$a_{5} = 2 + 4 \cdot 8 = 34$

✅ n = 5,

$a_n=34$

$a_{n} = 34$

(vii) a = 8,

$a_n=62$

$a_{n} = 62$ ,

$S_n=210$

$S_{n} = 210$ ; find n and d

$210=\frac{n}{2}(8+62)=\frac{n}{2}\cdot 70 \Rightarrow n=6$

$210 = 2 n (8 + 62) = 2 n \cdot 70 \Rightarrow n = 6$

$62=8+(6-1)d \Rightarrow 54=5d \Rightarrow d=\frac{54}{5}$

$62 = 8 + (6 - 1) d \Rightarrow 54 = 5 d \Rightarrow d = 554$

✅ n = 6, d = 54/5

(viii)

$a_n=4$

$a_{n} = 4$ , d = 2,

$S_n=-14$

$S_{n} = - 14$ ; find n and a

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

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

$-14=\frac{n}{2}(a+4)=\frac{n}{2}(6-2n+4)=\frac{n}{2}(10-2n)=n(5-n)$

$- 14 = 2 n (a + 4) = 2 n (6 - 2 n + 4) = 2 n (10 - 2 n) = n (5 - n)$

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

$n (5 - n) = - 14 \Rightarrow n^{2} - 5 n - 14 = 0 \Rightarrow (n - 7) (n + 2) = 0 \Rightarrow n = 7$

$a=6-14=-8$

$a = 6 - 14 = - 8$

✅ n = 7, a = −8

(ix) a = 3, n = 8,

$S_n=192$

$S_{n} = 192$ ; find d

$192=\frac{8}{2}[2\cdot 3+7d]=4(6+7d) \Rightarrow 48=6+7d \Rightarrow d=6$

$192 = 28 [2 \cdot 3 + 7 d] = 4 (6 + 7 d) \Rightarrow 48 = 6 + 7 d \Rightarrow d = 6$

✅ d = 6

(x) l = 28,

$S_9=144$

$S_{9} = 144$ ; find a

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

$144 = 29 (a + 28) \Rightarrow 288 = 9 (a + 28) \Rightarrow a + 28 = 32 \Rightarrow a = 4$

✅ a = 4

Q4. How many terms of the AP 9, 17, 25, … give sum 636?

a = 9, d = 8,

$S_n=636$

$S_{n} = 636$

$636=\frac{n}{2}[18+(n-1)8] \Rightarrow 1272=n[18+8n-8]=n(8n+10)$

$636 = 2 n [18 + (n - 1) 8] \Rightarrow 1272 = n [18 + 8 n - 8] = n (8 n + 10)$

$8n^2+10n-1272=0 \Rightarrow 4n^2+5n-636=0 \Rightarrow (n-12)(4n+53)=0 \Rightarrow n=12$

$8 n^{2} + 10 n - 1272 = 0 \Rightarrow 4 n^{2} + 5 n - 636 = 0 \Rightarrow (n - 12) (4 n + 53) = 0 \Rightarrow n = 12$

✅ n = 12

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

$400=\frac{n}{2}(5+45)=\frac{n}{2}\cdot 50=25n \Rightarrow n=16$

$400 = 2 n (5 + 45) = 2 n \cdot 50 = 25 n \Rightarrow n = 16$

$45=5+(16-1)d \Rightarrow 40=15d \Rightarrow d=\frac{8}{3}$

$45 = 5 + (16 - 1) d \Rightarrow 40 = 15 d \Rightarrow d = 38$

✅ n = 16, d = 8/3

Q6. First term 17, last term 350, d = 9. Find n and sum.

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

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

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

$S = 238 (17 + 350) = 19 \cdot 367 = 6973$

✅ n = 38, Sum = 6973

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

$149=a+21\cdot 7 \Rightarrow a=149-147=2$

$149 = a + 21 \cdot 7 \Rightarrow a = 149 - 147 = 2$

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

$S_{22} = 222 (2 + 149) = 11 \cdot 151 = 1661$

✅ Sum = 1661

Q8. Second term 14 and third term 18. Find sum of first 51 terms.

$d=18-14=4,\quad a=14-4=10$

$d = 18 - 14 = 4, a = 14 - 4 = 10$

$S_{51}=\frac{51}{2}[2\cdot 10+50\cdot 4] =\frac{51}{2}(20+200)=\frac{51}{2}\cdot 220=5610$

$S_{51} = 251 [2 \cdot 10 + 50 \cdot 4] = 251 (20 + 200) = 251 \cdot 220 = 5610$

✅ Sum = 5610

Q9. $S_7=49$

$S_{7} = 49$ and $S_{17}=289$

$S_{17} = 289$ . Find $S_n$

$S_{n}$ .

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

$S_{n} = 2 n [2 a + (n - 1) d]$

From

$S_7=49$

$S_{7} = 49$ :

$49=\frac{7}{2}(2a+6d)\Rightarrow 98=7(2a+6d)\Rightarrow 2a+6d=14 ...(1)$

$49 = 27 (2 a + 6 d) \Rightarrow 98 = 7 (2 a + 6 d) \Rightarrow 2 a + 6 d = 14... (1)$

From

$S_{17}=289$

$S_{17} = 289$ :

$289=\frac{17}{2}(2a+16d)\Rightarrow 578=17(2a+16d)\Rightarrow 2a+16d=34 ...(2)$

$289 = 217 (2 a + 16 d) \Rightarrow 578 = 17 (2 a + 16 d) \Rightarrow 2 a + 16 d = 34... (2)$

Subtract (1) from (2): 10d=20 ⇒ d=2
Then 2a+12=14 ⇒ a=1
So,

$S_n=\frac{n}{2}[2\cdot1+(n-1)2]=\frac{n}{2}[2+2n-2]=\frac{n}{2}(2n)=n^2$

$S_{n} = 2 n [2 \cdot 1 + (n - 1) 2] = 2 n [2 + 2 n - 2] = 2 n (2 n) = n^{2}$

✅

$S_n=n^2$

$S_{n} = n^{2}$

Q10. Show sequences form an AP and find sum of first 15 terms.

(i)

$a_n=3n+4$

$a_{n} = 3 n + 4$
a₁ = 7, common difference d = 4 ⇒ AP

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

$S_{15} = 215 [2 \cdot 7 + 14 \cdot 4] = 215 (14 + 56) = 215 \cdot 70 = 525$

✅ Sum = 525

(ii)

$a_n=9-5n$

$a_{n} = 9 - 5 n$
a₁ = 4, common difference d = −5 ⇒ AP

$S_{15}=\frac{15}{2}[2\cdot 4+14(-5)] =\frac{15}{2}(8-70)=\frac{15}{2}(-62)=-465$

$S_{15} = 215 [2 \cdot 4 + 14 (- 5)] = 215 (8 - 70) = 215 (- 62) = - 465$

✅ Sum = −465

Q11. If $S_n=2n-4n^2$

$S_{n} = 2 n - 4 n^{2}$ , find $S_1$

$S_{1}$ , $S_2$

$S_{2}$ , $a_2$

$a_{2}$ , $a_3$

$a_{3}$ , $a_{10}$

$a_{10}$ , $a_n$

$a_{n}$ .

$S_1=2(1)-4(1)^2=2-4=-2 \Rightarrow a_1=-2$

$S_{1} = 2 (1) - 4 (1)^{2} = 2 - 4 = - 2 \Rightarrow a_{1} = - 2$

$S_2=2(2)-4(4)=4-16=-12 \Rightarrow a_2=S_2-S_1=-12-(-2)=-10$

$S_{2} = 2 (2) - 4 (4) = 4 - 16 = - 12 \Rightarrow a_{2} = S_{2} - S_{1} = - 12 - (- 2) = - 10$

$S_3=2(3)-4(9)=6-36=-30 \Rightarrow a_3=S_3-S_2=-30-(-12)=-18$

$S_{3} = 2 (3) - 4 (9) = 6 - 36 = - 30 \Rightarrow a_{3} = S_{3} - S_{2} = - 30 - (- 12) = - 18$

$S_{10}=2(10)-4(100)=20-400=-380 \Rightarrow a_{10}=S_{10}-S_9$

$S_{10} = 2 (10) - 4 (100) = 20 - 400 = - 380 \Rightarrow a_{10} = S_{10} - S_{9}$

$S_9=2(9)-4(81)=18-324=-306 \Rightarrow a_{10}=-380-(-306)=-74$

$S_{9} = 2 (9) - 4 (81) = 18 - 324 = - 306 \Rightarrow a_{10} = - 380 - (- 306) = - 74$

General term:

$a_n=S_n-S_{n-1}=(2n-4n^2)-[2(n-1)-4(n-1)^2]=-8n+2$

$a_{n} = S_{n} - S_{n - 1} = (2 n - 4 n^{2}) - [2 (n - 1) - 4 (n - 1)^{2}] = - 8 n + 2$

✅

$S_1=-2,\; S_2=-12,\; a_2=-10,\; a_3=-18,\; a_{10}=-74,\; a_n=2-8n$

$S_{1} = - 2, S_{2} = - 12, a_{2} = - 10, a_{3} = - 18, a_{10} = - 74, a_{n} = 2 - 8 n$

Q12. Sum of first 40 positive integers divisible by 6

AP: 6, 12, 18, … (n=40)
a=6, d=6

$S_{40}=\frac{40}{2}[2\cdot6+39\cdot6] =20(12+234)=20\cdot246=4920$

$S_{40} = 240 [2 \cdot 6 + 39 \cdot 6] = 20 (12 + 234) = 20 \cdot 246 = 4920$

✅ Sum = 4920

Q13. Sum of first 15 multiples of 8

AP: 8, 16, 24, … (n=15)
a=8, d=8

$S_{15}=\frac{15}{2}[16+14\cdot 8] =\frac{15}{2}(16+112)=\frac{15}{2}\cdot128=960$

$S_{15} = 215 [16 + 14 \cdot 8] = 215 (16 + 112) = 215 \cdot 128 = 960$

✅ Sum = 960

Q14. Sum of odd numbers between 0 and 50

Odd numbers: 1, 3, 5, …, 49
a=1, d=2, l=49

$49=1+(n-1)2 \Rightarrow 48=2(n-1)\Rightarrow n=25$

$49 = 1 + (n - 1) 2 \Rightarrow 48 = 2 (n - 1) \Rightarrow n = 25$

$S=\frac{25}{2}(1+49)=\frac{25}{2}\cdot 50=625$

$S = 225 (1 + 49) = 225 \cdot 50 = 625$

✅ Sum = 625

Q15. Penalty AP: 200, 250, 300, … (30 days). Find total penalty.

a=200, d=50, n=30

$S_{30}=\frac{30}{2}[2\cdot200+29\cdot50] =15(400+1450)=15\cdot1850=27750$

$S_{30} = 230 [2 \cdot 200 + 29 \cdot 50] = 15 (400 + 1450) = 15 \cdot 1850 = 27750$

✅ Penalty = ₹27750

Q16. Total ₹700, 7 prizes, each ₹20 less than previous. Find prizes.

Let first prize = x
Prizes: x, x−20, x−40, … (7 terms), d=−20

$700=\frac{7}{2}[2x+6(-20)] =\frac{7}{2}(2x-120) \Rightarrow 1400=7(2x-120) \Rightarrow 200=2x-120 \Rightarrow x=160$

$700 = 27 [2 x + 6 (- 20)] = 27 (2 x - 120) \Rightarrow 1400 = 7 (2 x - 120) \Rightarrow 200 = 2 x - 120 \Rightarrow x = 160$

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

Q17. Tree planting (3 sections of each class I to XII). Find total trees.

Totals per class: 3, 6, 9, …, 36 (n=12)
a=3, d=3

$S_{12}=\frac{12}{2}[2\cdot 3+11\cdot 3] =6(6+33)=6\cdot 39=234$

$S_{12} = 212 [2 \cdot 3 + 11 \cdot 3] = 6 (6 + 33) = 6 \cdot 39 = 234$

✅ Total trees = 234

Q18. Spiral of 13 semicircles, radii: 0.5, 1.0, 1.5, … (take $\pi=\frac{22}{7}$

$π = 7 22$ ). Find total length.

Length of a semicircle =

$\pi r$

$π r$
So lengths form AP:

$0.5\pi, 1\pi, 1.5\pi, …$

$0.5 π, 1 π, 1.5 π, \dots$
a = 0.5π, d = 0.5π, n = 13

$S_{13}=\frac{13}{2}[2(0.5\pi)+12(0.5\pi)] =\frac{13}{2}[ \pi+6\pi ]=\frac{13}{2}\cdot 7\pi=\frac{91\pi}{2}$

$S_{13} = 213 [2 (0.5 π) + 12 (0.5 π)] = 213 [π + 6 π] = 213 \cdot 7 π = 291 π$

With

$\pi=\frac{22}{7}$

$π = 7 22$ :

$S=\frac{91}{2}\cdot \frac{22}{7}=143$

$S = 291 \cdot 722 = 143$

✅ Total length = 143 cm

Q19. 200 logs stacked: 20, 19, 18, … Find rows and logs in top row.

AP: a=20, d=−1,

$S_n=200$

$S_{n} = 200$

$200=\frac{n}{2}[2\cdot 20+(n-1)(-1)] =\frac{n}{2}[40-(n-1)] =\frac{n}{2}(41-n)$

$200 = 2 n [2 \cdot 20 + (n - 1) (- 1)] = 2 n [40 - (n - 1)] = 2 n (41 - n)$

$400=n(41-n)\Rightarrow n^2-41n+400=0 \Rightarrow (n-16)(n-25)=0 \Rightarrow n=16 \;(\text{valid})$

$400 = n (41 - n) \Rightarrow n^{2} - 41 n + 400 = 0 \Rightarrow (n - 16) (n - 25) = 0 \Rightarrow n = 16 (valid)$

Top row logs:

$a_{16}=20+15(-1)=5$

$a_{16} = 20 + 15 (- 1) = 5$

✅ Rows = 16, Top row = 5 logs

Q20. Potato race (10 potatoes): first at 5 m, others 3 m apart. Total distance run?

Distances from bucket: 5, 8, 11, 14, … (n=10)
a=5, d=3

$\text{Sum of distances}=\frac{10}{2}[2\cdot 5+9\cdot 3] =5(10+27)=185$

$Sum of distances = 210 [2 \cdot 5 + 9 \cdot 3] = 5 (10 + 27) = 185$

Each potato needs go + return, so multiply by 2:
✅ Total distance = 2 × 185 = 370 m

FAQs: Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

Q1. What is the focus of Exercise 5.3?
Answer:
Exercise 5.3 focuses on finding the nth term of an Arithmetic Progression (AP) and solving related problems.

Q2. What is the formula for the nth term of an AP?
Answer:
The nth term of an AP is given by:

$a_n = a + (n - 1)d$

$a_{n} = a + (n - 1) d$

where:

$a$ $a$ = first term
$d$ $d$ = common difference
$n$ $n$ = term number

Q3. How do I check if a number is a term of a given AP?
Answer:
Substitute the given number as

$a_n$

$a_{n}$ in the formula

$a_n = a + (n - 1)d$

$a_{n} = a + (n - 1) d$

and check whether you get a natural number value of

$n$

$n$ . If yes, it is a term of the AP.

Q4. What is the common difference in an AP?
Answer:
The common difference (d) is the difference between two consecutive terms:

$d = a_2 - a_1$

$d = a_{2} - a_{1}$

Q5. How do NCERT Solutions help in exam preparation?
Answer:
These solutions provide step-by-step explanations and help students understand how to apply the nth term formula correctly. Regular practice improves speed, accuracy, and confidence for board exams.

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

Q1. Find the sum of the following APs.

Q2. Find the sums given below

Q3. In an AP

Q4. How many terms of the AP 9, 17, 25, … give sum 636?

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

Q6. First term 17, last term 350, d = 9. Find n and sum.

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

Q8. Second term 14 and third term 18. Find sum of first 51 terms.

Q9. $S_7=49$

$S_{7} = 49$ and $S_{17}=289$

$S_{17} = 289$ . Find $S_n$

$S_{n}$ .

Q10. Show sequences form an AP and find sum of first 15 terms.

Q11. If $S_n=2n-4n^2$

$S_{n} = 2 n - 4 n^{2}$ , find $S_1$

$S_{1}$ , $S_2$

$S_{2}$ , $a_2$

$a_{2}$ , $a_3$

$a_{3}$ , $a_{10}$

$a_{10}$ , $a_n$

$a_{n}$ .

Q12. Sum of first 40 positive integers divisible by 6

Q13. Sum of first 15 multiples of 8

Q14. Sum of odd numbers between 0 and 50

Q15. Penalty AP: 200, 250, 300, … (30 days). Find total penalty.

Q16. Total ₹700, 7 prizes, each ₹20 less than previous. Find prizes.

Q17. Tree planting (3 sections of each class I to XII). Find total trees.

Q18. Spiral of 13 semicircles, radii: 0.5, 1.0, 1.5, … (take $\pi=\frac{22}{7}$

$π = 7 22$ ). Find total length.

Q19. 200 logs stacked: 20, 19, 18, … Find rows and logs in top row.

Q20. Potato race (10 potatoes): first at 5 m, others 3 m apart. Total distance run?

FAQs: Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

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NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

Q1. Find the sum of the following APs.

Q2. Find the sums given below

Q3. In an AP

Q4. How many terms of the AP 9, 17, 25, … give sum 636?

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

Q6. First term 17, last term 350, d = 9. Find n and sum.

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

Q8. Second term 14 and third term 18. Find sum of first 51 terms.

Q9. S7=49S_7=49

S7​=49 and S17=289S_{17}=289

S17​=289. Find SnS_n

Sn​.

Q10. Show sequences form an AP and find sum of first 15 terms.

Q11. If Sn=2n−4n2S_n=2n-4n^2

Sn​=2n−4n2, find S1S_1

S1​, S2S_2

S2​, a2a_2

a2​, a3a_3

a3​, a10a_{10}

a10​, ana_n

an​.

Q12. Sum of first 40 positive integers divisible by 6

Q13. Sum of first 15 multiples of 8

Q14. Sum of odd numbers between 0 and 50

Q15. Penalty AP: 200, 250, 300, … (30 days). Find total penalty.

Q16. Total ₹700, 7 prizes, each ₹20 less than previous. Find prizes.

Q17. Tree planting (3 sections of each class I to XII). Find total trees.

Q18. Spiral of 13 semicircles, radii: 0.5, 1.0, 1.5, … (take π=227\pi=\frac{22}{7}

π=722​). Find total length.

Q19. 200 logs stacked: 20, 19, 18, … Find rows and logs in top row.

Q20. Potato race (10 potatoes): first at 5 m, others 3 m apart. Total distance run?

FAQs: Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

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Q9. $S_7=49$

$S_{7} = 49$ and $S_{17}=289$

$S_{17} = 289$ . Find $S_n$

$S_{n}$ .

Q11. If $S_n=2n-4n^2$

$S_{n} = 2 n - 4 n^{2}$ , find $S_1$

$S_{1}$ , $S_2$

$S_{2}$ , $a_2$

$a_{2}$ , $a_3$

$a_{3}$ , $a_{10}$

$a_{10}$ , $a_n$

$a_{n}$ .

Q18. Spiral of 13 semicircles, radii: 0.5, 1.0, 1.5, … (take $\pi=\frac{22}{7}$

$π = 7 22$ ). Find total length.