NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (NCERT 2026–27)

These Class 10 Maths Chapter 5 solutions cover Arithmetic Progressions from the NCERT textbook (Reprint 2026–27). Every question of Exercise 5.1, 5.2, 5.3 and the optional Exercise 5.4 is solved step by step using the nth-term formula an = a + (n − 1)d and the sum formula Sn = (n/2)[2a + (n − 1)d], with every answer cross-checked against the book.

Class: 10 Subject: Mathematics Chapter: 5 Chapter Name: Arithmetic Progressions Exercises: 5.1, 5.2, 5.3, 5.4 (Optional) Session: 2026–27
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Chapter 5 Overview

Chapter 5, Arithmetic Progressions, studies lists of numbers in which each term is obtained by adding a fixed number, called the common difference (d), to the previous term. The chapter starts with everyday patterns (salaries with annual increments, ladder rungs, savings), defines the general form a, a + d, a + 2d, …, then derives two key tools: the nth term an = a + (n − 1)d and the sum of the first n terms Sn. Using these, you learn to find any term, the number of terms, missing terms, and the sum of a finite AP, and to apply them to real-life word problems. The Class 10 Maths Chapter 5 solutions below work through every exercise question with full steps.

Key Concepts & Definitions

Arithmetic Progression (AP): a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.

Common difference (d): the fixed number added each time, d = ak+1 − ak. It can be positive, negative or zero.

First term (a): the starting term a1. The general form of an AP is a, a + d, a + 2d, a + 3d, …

Finite vs infinite AP: an AP with a last term is finite; one without a last term is infinite.

nth term / general term (an): the term at position n, given by an = a + (n − 1)d.

Last term (l): if there are n terms, l = an = a + (n − 1)d.

Arithmetic mean: if a, b, c are in AP, then b = (a + c)/2 is the arithmetic mean of a and c.

Important Formulas (Chapter 5)

nth term: an = a + (n − 1)d.

Common difference: d = an − an−1 (subtract any term from the term after it).

Sum of first n terms: Sn = (n/2)[2a + (n − 1)d].

Sum using first & last term: Sn = (n/2)(a + l), where l = an is the last term.

Relation between term and sum: an = Sn − Sn−1.

Sum of first n natural numbers: 1 + 2 + 3 + … + n = n(n + 1)/2.

Exercise 5.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the book’s answers.

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre. (iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.

SOLUTION (i) Fares: 15, 23, 31, 39, … Each term increases by ₹ 8. Common difference is constant → it is an AP (d = 8). (ii) Each step leaves 3/4 of the previous amount, so the terms are multiplied (not added) by a fixed number. The difference is not constant → not an AP. (iii) Costs: 150, 200, 250, 300, … Each term increases by ₹ 50 → it is an AP (d = 50). (iv) With compound interest the amount is multiplied by 1.08 each year, so successive differences keep increasing → not an AP.

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10    (ii) a = −2, d = 0 (iii) a = 4, d = −3    (iv) a = −1, d = 1/2 (v) a = −1.25, d = −0.25

SOLUTION Use a, a + d, a + 2d, a + 3d. (i) 10, 20, 30, 40. (ii) −2, −2, −2, −2. (iii) 4, 1, −2, −5. (iv) −1, −1/2, 0, 1/2. (v) −1.25, −1.50, −1.75, −2.00.

3. For the following APs, write the first term and the common difference: (i) 3, 1, −1, −3, . . .    (ii) −5, −1, 3, 7, . . . (iii) 1/3, 5/3, 9/3, 13/3, . . .    (iv) 0.6, 1.7, 2.8, 3.9, . . .

SOLUTION First term a = a1; common difference d = a2 − a1. (i) a = 3, d = 1 − 3 = −2. (ii) a = −5, d = −1 − (−5) = 4. (iii) a = 1/3, d = 5/3 − 1/3 = 4/3. (iv) a = 0.6, d = 1.7 − 0.6 = 1.1.

4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16, . . .    (ii) 2, 5/2, 3, 7/2, . . . (iii) −1.2, −3.2, −5.2, −7.2, . . .    (iv) −10, −6, −2, 2, . . . (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, . . .    (vi) 0.2, 0.22, 0.222, 0.2222, . . . (vii) 0, −4, −8, −12, . . .    (viii) −1/2, −1/2, −1/2, −1/2, . . . (ix) 1, 3, 9, 27, . . .    (x) a, 2a, 3a, 4a, . . . (xi) a, a2, a3, a4, . . .    (xii) √2, √8, √18, √32, . . . (xiii) √3, √6, √9, √12, . . .    (xiv) 12, 32, 52, 72, . . . (xv) 12, 52, 72, 73, . . .

SOLUTION (i) 4 − 2 = 2, 8 − 4 = 4. Differences differ → not an AP. (ii) 5/2 − 2 = 1/2, 3 − 5/2 = 1/2. AP, d = 1/2; next three: 4, 9/2, 5. (iii) Each difference = −2. AP, d = −2; next three: −9.2, −11.2, −13.2. (iv) Each difference = 4. AP, d = 4; next three: 6, 10, 14. (v) Each difference = √2. AP, d = √2; next three: 3 + 4√2, 3 + 5√2, 3 + 6√2. (vi) 0.22 − 0.2 = 0.02, 0.222 − 0.22 = 0.002. Differences differ → not an AP. (vii) Each difference = −4. AP, d = −4; next three: −16, −20, −24. (viii) Each difference = 0. AP, d = 0; next three: −1/2, −1/2, −1/2. (ix) 3 − 1 = 2, 9 − 3 = 6. Differences differ → not an AP. (x) Each difference = a. AP, d = a; next three: 5a, 6a, 7a. (xi) a2 − a = a(a − 1), a3 − a2 = a2(a − 1); not equal in general → not an AP. (xii) √2, 2√2, 3√2, 4√2 (since √8 = 2√2, √18 = 3√2, √32 = 4√2). Each difference = √2. AP, d = √2; next three: 5√2 (= √50), 6√2 (= √72), 7√2 (= √98). (xiii) √3, √6, √9, √12 = √3, √6, 3, 2√3. √6 − √3 ≠ 3 − √6 → not an AP. (xiv) 1, 9, 25, 49. 9 − 1 = 8, 25 − 9 = 16. Differences differ → not an AP. (xv) 1, 25, 49, 73. 25 − 1 = 24, 49 − 25 = 24, 73 − 49 = 24. Each difference = 24. AP, d = 24; next three: 97, 121, 145.

Exercise 5.2 Solutions

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

adnan
(i)738…
(ii)−18…100
(iii)…−318−5
(iv)−18.92.5…3.6
(v)3.50105…
SOLUTION Use an = a + (n − 1)d. (i) a8 = 7 + 7×3 = 7 + 21 = 28. (ii) 0 = −18 + 9d ⇒ 9d = 18 ⇒ d = 2. (iii) −5 = a + 17(−3) = a − 51 ⇒ a = 46. (iv) 3.6 = −18.9 + (n − 1)(2.5) ⇒ 22.5 = 2.5(n − 1) ⇒ n − 1 = 9 ⇒ n = 10. (v) a105 = 3.5 + 104×0 = 3.5.

2. 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, −1/2, 2, . . ., is   (A) 28   (B) 22   (C) −38   (D) −48½

SOLUTION (i) a = 10, d = 7 − 10 = −3. a30 = 10 + 29(−3) = 10 − 87 = −77. Correct option (C) −77. (ii) a = −3, d = −1/2 − (−3) = 5/2. a11 = −3 + 10(5/2) = −3 + 25 = 22. Correct option (B) 22.

3. In the following APs, find the missing terms in the boxes: (i) 2, __, 26 (ii) __, 13, __, 3 (iii) 5, __, __, 9½ (iv) −4, __, __, __, __, 6 (v) __, 38, __, __, __, −22

SOLUTION (i) Three terms 2, x, 26 in AP ⇒ x is the middle term = (2 + 26)/2 = 14. Missing term = 14. (ii) Let terms be a1, 13, a3, 3. Here a2 = 13 and a4 = 3, so 13 = a + d and 3 = a + 3d. Subtract: 2d = −10 ⇒ d = −5, a = 18. Terms: 18, 13, 8, 3; missing are 18 and 8. (iii) Terms 5, x, y, 9½ (i.e. 19/2). a = 5, a4 = 5 + 3d = 19/2 ⇒ 3d = 9/2 ⇒ d = 3/2. So x = 5 + 3/2 = 13/2 (6½), y = 5 + 3 = 8. (iv) Terms −4, …, 6 with 6 terms. a = −4, a6 = −4 + 5d = 6 ⇒ 5d = 10 ⇒ d = 2. Terms: −4, −2, 0, 2, 4, 6. (v) Terms with a2 = 38 and a6 = −22. a6 − a2 = 4d = −60 ⇒ d = −15; a = 38 − d = 53. Terms: 53, 38, 23, 8, −7, −22.

4. Which term of the AP: 3, 8, 13, 18, . . . , is 78?

SOLUTION a = 3, d = 5. 78 = 3 + (n − 1)5 ⇒ 75 = 5(n − 1) ⇒ n − 1 = 15 ⇒ n = 16. So 78 is the 16th term.

5. Find the number of terms in each of the following APs: (i) 7, 13, 19, . . . , 205    (ii) 18, 15½, 13, . . . , −47

SOLUTION (i) a = 7, d = 6, an = 205. 205 = 7 + (n − 1)6 ⇒ 198 = 6(n − 1) ⇒ n − 1 = 33 ⇒ n = 34. (ii) a = 18, d = 15½ − 18 = −5/2, an = −47. −47 = 18 + (n − 1)(−5/2) ⇒ −65 = (n − 1)(−5/2) ⇒ n − 1 = 26 ⇒ n = 27.

6. Check whether −150 is a term of the AP: 11, 8, 5, 2 . . .

SOLUTION a = 11, d = −3. Suppose an = −150: −150 = 11 + (n − 1)(−3) ⇒ −161 = −3(n − 1) ⇒ n − 1 = 161/3, which is not a whole number. Since n is not a positive integer, −150 is not a term of this AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

SOLUTION a + 10d = 38 and a + 15d = 73. Subtract: 5d = 35 ⇒ d = 7; a = 38 − 70 = −32. a31 = −32 + 30×7 = −32 + 210 = 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

SOLUTION a + 2d = 12 and a50 = a + 49d = 106. Subtract: 47d = 94 ⇒ d = 2; a = 12 − 4 = 8. a29 = 8 + 28×2 = 8 + 56 = 64.

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

SOLUTION a + 2d = 4 and a + 8d = −8. Subtract: 6d = −12 ⇒ d = −2; a = 4 + 4 = 8. an = 0: 8 + (n − 1)(−2) = 0 ⇒ (n − 1) = 4 ⇒ n = 5. So the 5th term is zero.

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

SOLUTION a17 − a10 = 7 ⇒ (a + 16d) − (a + 9d) = 7 ⇒ 7d = 7 ⇒ d = 1.

11. Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54th term?

SOLUTION a = 3, d = 12. a54 = 3 + 53×12 = 639. Required term = 639 + 132 = 771. 771 = 3 + (n − 1)12 ⇒ 768 = 12(n − 1) ⇒ n − 1 = 64 ⇒ n = 65. So the 65th term is 132 more than the 54th.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

SOLUTION Let first terms be a and A, common difference d each. 100th terms: a + 99d and A + 99d; their difference = a − A = 100. 1000th terms: a + 999d and A + 999d; their difference = a − A = 100 (the difference equals a − A for any term).

13. How many three-digit numbers are divisible by 7?

SOLUTION Three-digit multiples of 7: 105, 112, …, 994. a = 105, d = 7, an = 994. 994 = 105 + (n − 1)7 ⇒ 889 = 7(n − 1) ⇒ n − 1 = 127 ⇒ n = 128. So there are 128 three-digit numbers divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

SOLUTION Multiples of 4 between 10 and 250: 12, 16, …, 248. a = 12, d = 4, an = 248. 248 = 12 + (n − 1)4 ⇒ 236 = 4(n − 1) ⇒ n − 1 = 59 ⇒ n = 60. So there are 60 multiples of 4.

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

SOLUTION First AP: a = 63, d = 2 ⇒ an = 63 + (n − 1)2 = 61 + 2n. Second AP: a = 3, d = 7 ⇒ an = 3 + (n − 1)7 = 7n − 4. Set equal: 61 + 2n = 7n − 4 ⇒ 65 = 5n ⇒ n = 13.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

SOLUTION a + 2d = 16. Also a7 − a5 = 12 ⇒ (a + 6d) − (a + 4d) = 12 ⇒ 2d = 12 ⇒ d = 6. Then a = 16 − 12 = 4. The AP is 4, 10, 16, 22, . . .

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

SOLUTION Counting from the last, treat 253 as the first term with d reversed to −5. The 20th term from the last = 253 + (20 − 1)(−5) = 253 − 95 = 158.

18. 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.

SOLUTION a4 + a8 = (a + 3d) + (a + 7d) = 2a + 10d = 24 ⇒ a + 5d = 12. a6 + a10 = (a + 5d) + (a + 9d) = 2a + 14d = 44 ⇒ a + 7d = 22. Subtract: 2d = 10 ⇒ d = 5; a = 12 − 25 = −13. First three terms: −13, −8, −3.

19. 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?

SOLUTION Salaries form an AP: a = 5000, d = 200. Let an = 7000: 7000 = 5000 + (n − 1)200 ⇒ 2000 = 200(n − 1) ⇒ n − 1 = 10 ⇒ n = 11. The 11th year (counting 1995 as year 1) is 1995 + 10 = 2005.

20. 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.

SOLUTION a = 5, d = 1.75, an = 20.75. 20.75 = 5 + (n − 1)(1.75) ⇒ 15.75 = 1.75(n − 1) ⇒ n − 1 = 9 ⇒ n = 10.

Exercise 5.3 Solutions

1. 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) 1/15, 1/12, 1/10, . . ., to 11 terms.

SOLUTION Use Sn = (n/2)[2a + (n − 1)d]. (i) a = 2, d = 5, n = 10: S = 5[4 + 9×5] = 5×49 = 245. (ii) a = −37, d = 4, n = 12: S = 6[−74 + 11×4] = 6[−74 + 44] = 6(−30) = −180. (iii) a = 0.6, d = 1.1, n = 100: S = 50[1.2 + 99×1.1] = 50[1.2 + 108.9] = 50×110.1 = 5505. (iv) a = 1/15, d = 1/12 − 1/15 = 1/60, n = 11: S = (11/2)[2/15 + 10/60] = (11/2)[2/15 + 1/6] = (11/2)(9/30) = (11/2)(3/10) = 33/20 (= 1.65).

2. Find the sums given below: (i) 7 + 10½ + 14 + . . . + 84    (ii) 34 + 32 + 30 + . . . + 10 (iii) −5 + (−8) + (−11) + . . . + (−230)

SOLUTION (i) a = 7, d = 7/2, l = 84. 84 = 7 + (n − 1)(7/2) ⇒ 77 = (7/2)(n − 1) ⇒ n − 1 = 22 ⇒ n = 23. S = (23/2)(7 + 84) = (23/2)(91) = 1046½ (2093/2). (ii) a = 34, d = −2, l = 10. 10 = 34 + (n − 1)(−2) ⇒ −24 = −2(n − 1) ⇒ n − 1 = 12 ⇒ n = 13. S = (13/2)(34 + 10) = (13/2)(44) = 286. (iii) a = −5, d = −3, l = −230. −230 = −5 + (n − 1)(−3) ⇒ −225 = −3(n − 1) ⇒ n − 1 = 75 ⇒ n = 76. S = (76/2)(−5 − 230) = 38×(−235) = −8930.

3. In an AP: (i) given a = 5, d = 3, an = 50, find n and Sn. (ii) given a = 7, a13 = 35, find d and S13. (iii) given a12 = 37, d = 3, find a and S12. (iv) given a3 = 15, S10 = 125, find d and a10. (v) given d = 5, S9 = 75, find a and a9. (vi) given a = 2, d = 8, Sn = 90, find n and an. (vii) given a = 8, an = 62, Sn = 210, find n and d. (viii) given an = 4, d = 2, Sn = −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.

SOLUTION (i) 50 = 5 + (n − 1)3 ⇒ 45 = 3(n − 1) ⇒ n = 16; S16 = (16/2)(5 + 50) = 8×55 = 440. (ii) a13 = 7 + 12d = 35 ⇒ 12d = 28 ⇒ d = 7/3; S13 = (13/2)(7 + 35) = (13/2)(42) = 273. (iii) a12 = a + 11×3 = 37 ⇒ a = 37 − 33 = 4; S12 = (12/2)(4 + 37) = 6×41 = 246. (iv) a + 2d = 15 and S10 = (10/2)[2a + 9d] = 5(2a + 9d) = 125 ⇒ 2a + 9d = 25. From a = 15 − 2d: 2(15 − 2d) + 9d = 25 ⇒ 30 + 5d = 25 ⇒ d = −1; a = 17. a10 = 17 + 9(−1) = 8. (v) S9 = (9/2)[2a + 8d] = (9/2)(2a + 40) = 75 ⇒ 2a + 40 = 50/3 ⇒ 2a = 50/3 − 40 = −70/3 ⇒ a = −35/3; a9 = a + 8d = −35/3 + 40 = 85/3. (vi) Sn = (n/2)[4 + (n − 1)8] = (n/2)(8n − 4) = 4n2 − 2n = 90 ⇒ 4n2 − 2n − 90 = 0 ⇒ 2n2 − n − 45 = 0 ⇒ (2n + 9)(n − 5) = 0 ⇒ n = 5; a5 = 2 + 4×8 = 34. (vii) Sn = (n/2)(a + an) = (n/2)(8 + 62) = 35n = 210 ⇒ n = 6; an = 8 + 5d = 62 ⇒ d = 54/5 (= 10.8). (viii) an = a + (n − 1)2 = 4. Sn = (n/2)(a + 4) = −14. From the first: a = 4 − 2(n − 1) = 6 − 2n. Substitute: (n/2)(10 − 2n) = −14 ⇒ n(5 − n) = −14 ⇒ n2 − 5n − 14 = 0 ⇒ (n − 7)(n + 2) = 0 ⇒ n = 7; a = 6 − 14 = −8. (ix) S8 = (8/2)[2×3 + 7d] = 4(6 + 7d) = 192 ⇒ 6 + 7d = 48 ⇒ d = 6. (x) S = (9/2)(a + 28) = 144 ⇒ a + 28 = 32 ⇒ a = 4.

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

SOLUTION a = 9, d = 8. Sn = (n/2)[18 + (n − 1)8] = (n/2)(8n + 10) = 4n2 + 5n = 636. 4n2 + 5n − 636 = 0 ⇒ n = [−5 ± √(25 + 10176)]/8 = [−5 ± 101]/8. Taking the positive root, n = 96/8 = 12.

5. 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 S = (n/2)(a + l): 400 = (n/2)(5 + 45) = 25n ⇒ n = 16. l = a + (n − 1)d: 45 = 5 + 15d ⇒ 15d = 40 ⇒ d = 8/3.

6. 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 l = a + (n − 1)d: 350 = 17 + (n − 1)9 ⇒ 333 = 9(n − 1) ⇒ n − 1 = 37 ⇒ n = 38. S = (38/2)(17 + 350) = 19×367 = 6973.

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

SOLUTION a22 = a + 21×7 = 149 ⇒ a = 149 − 147 = 2. S22 = (22/2)(a + a22) = 11(2 + 149) = 11×151 = 1661.

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

SOLUTION d = 18 − 14 = 4; a = a2 − d = 14 − 4 = 10. S51 = (51/2)[2×10 + 50×4] = (51/2)(20 + 200) = (51/2)(220) = 51×110 = 5610.

9. 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 S7 = (7/2)(2a + 6d) = 49 ⇒ 2a + 6d = 14 ⇒ a + 3d = 7. S17 = (17/2)(2a + 16d) = 289 ⇒ 2a + 16d = 34 ⇒ a + 8d = 17. Subtract: 5d = 10 ⇒ d = 2; a = 7 − 6 = 1. Sn = (n/2)[2 + (n − 1)2] = (n/2)(2n) = n2.

10. Show that a1, a2, . . ., an, . . . form an AP where an is defined as below: (i) an = 3 + 4n    (ii) an = 9 − 5n Also find the sum of the first 15 terms in each case.

SOLUTION (i) an+1 − an = [3 + 4(n + 1)] − [3 + 4n] = 4, a constant → it is an AP with a = a1 = 7, d = 4. S15 = (15/2)[14 + 14×4] = (15/2)(14 + 56) = (15/2)(70) = 525. (ii) an+1 − an = [9 − 5(n + 1)] − [9 − 5n] = −5, a constant → it is an AP with a = a1 = 4, d = −5. S15 = (15/2)[8 + 14(−5)] = (15/2)(8 − 70) = (15/2)(−62) = −465.

11. If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

SOLUTION S1 = 4(1) − 1 = 3, so first term a1 = 3. S2 = 4(2) − 4 = 4 (sum of first two terms). Second term a2 = S2 − S1 = 4 − 3 = 1. In general an = Sn − Sn−1 = (4n − n2) − [4(n − 1) − (n − 1)2] = 5 − 2n. So a3 = 5 − 6 = −1; a10 = 5 − 20 = −15; the nth term an = 5 − 2n.

12. Find the sum of the first 40 positive integers divisible by 6.

SOLUTION Multiples of 6: 6, 12, 18, … a = 6, d = 6, n = 40. S40 = (40/2)[12 + 39×6] = 20(12 + 234) = 20×246 = 4920.

13. Find the sum of the first 15 multiples of 8.

SOLUTION Multiples of 8: 8, 16, 24, … a = 8, d = 8, n = 15. S15 = (15/2)[16 + 14×8] = (15/2)(16 + 112) = (15/2)(128) = 15×64 = 960.

14. Find the sum of the odd numbers between 0 and 50.

SOLUTION Odd numbers: 1, 3, 5, …, 49. a = 1, d = 2, l = 49 ⇒ n = 25. S25 = (25/2)(1 + 49) = (25/2)(50) = 625.

15. 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 Penalties form an AP: a = 200, d = 50, n = 30. S30 = (30/2)[400 + 29×50] = 15(400 + 1450) = 15×1850 = ₹ 27750.

16. 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 Prizes form an AP with d = −20, n = 7, S7 = 700. S7 = (7/2)[2a + 6(−20)] = 700 ⇒ (7/2)(2a − 120) = 700 ⇒ 2a − 120 = 200 ⇒ a = 160. Prizes (in ₹): 160, 140, 120, 100, 80, 60, 40.

17. 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 Trees per class (one section): 1, 2, 3, …, 12. With three sections each, total = 3 × (1 + 2 + … + 12). 1 + 2 + … + 12 = (12×13)/2 = 78. Total = 3 × 78 = 234 trees.

18. 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, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

SOLUTION Length of a semicircle = πr. Radii (in cm): 0.5, 1.0, 1.5, …, i.e. rn = 0.5n = n/2 for n = 1 to 13. Total length = π(r1 + r2 + … + r13) = π × (1/2)(1 + 2 + … + 13) = π × (1/2)(91) = (22/7) × (91/2). = (22 × 91)/(7 × 2) = 2002/14 = 143 cm.

19. 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 Logs per row: 20, 19, 18, … a = 20, d = −1, Sn = 200. (n/2)[40 + (n − 1)(−1)] = 200 ⇒ (n/2)(41 − n) = 200 ⇒ n2 − 41n + 400 = 0 ⇒ (n − 16)(n − 25) = 0 ⇒ n = 16 or 25. For n = 25, top row count a25 = 20 + 24(−1) = −4, which is impossible. So n = 16 rows. Top row a16 = 20 + 15(−1) = 5 logs.

20. 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 For each potato the competitor runs to it and back, so the distance is twice the potato’s distance from the bucket. Distances to potatoes: 5, 8, 11, … (3 m apart), so distances run: 2×5, 2×8, 2×11, … = 10, 16, 22, … This is an AP with a = 10, d = 6, n = 10. S10 = (10/2)[20 + 9×6] = 5(20 + 54) = 5×74 = 370 m.

Exercise 5.4 (Optional) Solutions

These exercises are not from the examination point of view, but they make excellent higher-order practice.

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an < 0]

SOLUTION a = 121, d = −4. an = 121 + (n − 1)(−4) = 125 − 4n. For an < 0: 125 − 4n < 0 ⇒ n > 125/4 = 31.25. The smallest integer is n = 32. Check: a32 = 125 − 128 = −3 (< 0). So the 32nd term is the first negative term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

SOLUTION a3 + a7 = (a + 2d) + (a + 6d) = 2a + 8d = 6 ⇒ a + 4d = 3. Product: (a + 2d)(a + 6d) = 8. Writing a + 2d = 3 − 2d and a + 6d = 3 + 2d gives (3 − 2d)(3 + 2d) = 9 − 4d2 = 8 ⇒ 4d2 = 1 ⇒ d = ±1/2. Case d = 1/2: a = 3 − 2 = 1. S16 = (16/2)[2 + 15(1/2)] = 8(2 + 7.5) = 8×9.5 = 76. Case d = −1/2: a = 3 + 2 = 5. S16 = (16/2)[10 + 15(−1/2)] = 8(10 − 7.5) = 8×2.5 = 20. So the sum of the first sixteen terms is 76 or 20.

3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = 250/25 + 1]

SOLUTION Distance between top and bottom rungs = 2½ m = 250 cm; rungs are 25 cm apart. Number of rungs = 250/25 + 1 = 11. Rung lengths form an AP: a = 25, l = 45, n = 11 (or first 45, last 25 — the sum is the same). Total length of wood = S11 = (11/2)(25 + 45) = (11/2)(70) = 11×35 = 385 cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint: Sx−1 = S49 − Sx]

SOLUTION House numbers 1, 2, …, 49. Sum preceding x = 1 + 2 + … + (x − 1) = (x − 1)x/2. Sum following x = (x + 1) + … + 49 = S49 − Sx, where S49 = (49×50)/2 = 1225 and Sx = x(x + 1)/2. Set them equal: (x − 1)x/2 = 1225 − x(x + 1)/2 ⇒ (x2 − x) + (x2 + x) = 2450 ⇒ 2x2 = 2450 ⇒ x2 = 1225 ⇒ x = 35. Since 35 lies between 1 and 49, such a house exists and x = 35.

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = 1/4 × 1/2 × 50 m3]

SOLUTION Each step is 50 m long with tread 1/2 m. The first step is one block of height 1/4 m; the second step rests on two such heights (1/2 m), the third on three (3/4 m), and so on, so heights form an AP: 1/4, 2/4, 3/4, …, 15/4 m. Volume of the nth step = length × tread × height = 50 × (1/2) × (n/4) = 25n/4 m3. Total volume = (25/4)(1 + 2 + … + 15) = (25/4) × (15×16)/2 = (25/4) × 120 = 25 × 30 = 750 m3.

Common Mistakes to Avoid

Watch out for these

  • Using n instead of (n − 1) in an = a + (n − 1)d — the very first term needs no addition of d.
  • Finding d the wrong way round: always subtract a term from the term after it (a2 − a1), even when terms decrease.
  • Forgetting to check that n comes out a positive whole number; a fractional n means the value is not a term of the AP.
  • Confusing the two sum forms — use (n/2)(a + l) only when you already know the last term l.
  • In word problems (logs, rows), rejecting an answer that gives a negative or impossible term — only the physically valid value of n is acceptable.
  • Counting the “kth term from the last” as the kth from the start — reverse the AP (d becomes −d) or use n − k + 1.

Practice MCQs & Assertion–Reason

1. The common difference of the AP 5, 8, 11, 14, . . . is:

(a) 2    (b) 3    (c) −3    (d) 5

2. The 10th term of the AP 2, 7, 12, . . . is:

(a) 45    (b) 47    (c) 50    (d) 52

3. The nth term of an AP is given by an = 4n + 1. Its common difference is:

(a) 1    (b) 4    (c) 5    (d) 9

4. The sum of the first 10 natural numbers is:

(a) 45    (b) 50    (c) 55    (d) 100

5. Which term of the AP 3, 8, 13, . . . is 78?

(a) 14th    (b) 15th    (c) 16th    (d) 17th

6. If the first term is 5 and the last term is 45 of an AP of 16 terms, its sum is:

(a) 360    (b) 400    (c) 500    (d) 800

7. The number of three-digit numbers divisible by 7 is:

(a) 127    (b) 128    (c) 129    (d) 130

8. The sum Sn of the first n terms of an AP is given by Sn = n2. The nth term an is:

(a) n    (b) 2n    (c) 2n − 1    (d) n2 − 1

9. For an AP, a = 7, d = 3 and an = 28. The value of n is:

(a) 6    (b) 7    (c) 8    (d) 9

10. The 30th term of the AP 10, 7, 4, . . . is:

(a) 97    (b) 77    (c) −77    (d) −87

Answer key: 1-(b), 2-(b), 3-(b), 4-(c), 5-(c), 6-(b), 7-(b), 8-(c), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The list 2, 4, 8, 16, . . . is an AP.

Reason: In an AP the difference between consecutive terms is constant.

A-R 2. Assertion: The 11th term of the AP −3, −1/2, 2, . . . is 22.

Reason: The nth term of an AP is an = a + (n − 1)d.

A-R 3. Assertion: The sum of the first n positive integers is n(n + 1)/2.

Reason: 1, 2, 3, . . . , n is an AP with first term 1 and common difference 1.

A-R 4. Assertion: −150 is a term of the AP 11, 8, 5, 2, . . .

Reason: A number is a term of an AP only if the value of n obtained from an = a + (n − 1)d is a positive integer.

A-R 5. Assertion: If two APs have the same common difference, the difference between their corresponding terms stays constant.

Reason: The common difference cancels out when the nth terms of the two APs are subtracted.

Answer key: 1-(D), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • An AP is a list where each term differs from the previous by a fixed common difference d; general form a, a + d, a + 2d, …
  • A list is an AP iff ak+1 − ak is the same for all k.
  • nth term: an = a + (n − 1)d — use it to find any term, the number of terms, or to check if a value is a term.
  • Sum of n terms: Sn = (n/2)[2a + (n − 1)d], or Sn = (n/2)(a + l) when the last term l is known.
  • Relation: an = Sn − Sn−1.
  • If a, b, c are in AP then b = (a + c)/2, the arithmetic mean.

How to score full marks in this chapter

Always write down a and d first, then quote the correct formula (an for “which term/find a term” questions, Sn for “sum” questions). Substitute carefully, show each algebra step, and finish by stating the answer with units. In word problems, identify the AP clearly, reject any impossible value of n, and double-check arithmetic on n − 1 to avoid the most common slip.

Frequently Asked Questions

What is Class 10 Maths Chapter 5 Arithmetic Progressions about?

Chapter 5 studies arithmetic progressions — lists of numbers with a constant common difference. You learn the nth-term formula an = a + (n − 1)d and the sum formula Sn = (n/2)[2a + (n − 1)d], and apply them to find terms, the number of terms, missing terms, sums, and real-life problems.

How many exercises are there in Class 10 Maths Chapter 5?

There are four exercises — Exercise 5.1, 5.2 and 5.3, plus an optional Exercise 5.4 marked “not from the examination point of view.” All four are solved in full on this page.

What are the two main formulas in this chapter?

The nth term an = a + (n − 1)d and the sum of the first n terms Sn = (n/2)[2a + (n − 1)d] = (n/2)(a + l), where a is the first term, d the common difference and l the last term.

Are these Class 10 Maths Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Class 10 Mathematics textbook for the 2026–27 session, with every answer verified against the book.

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