Arithmetic Progressions is a large, formula-heavy chapter in the current 2026-27 CBSE Class 10 Maths syllabus, spread across three exercises (5.1, 5.2, 5.3) — the earlier optional Exercise 5.4 was removed in the NCERT rationalisation and is not part of the current syllabus. Below are complete, verified solutions to a representative core set of questions covering every question type across all three exercises.
NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions
Exercise 5.1
Q1(i). A taxi fare is Rs 15 for the first km and Rs 8 for each additional km. Is the fare for 1, 2, 3, … km in AP?
Fares: 15, 23, 31, 39, … Common difference = 8 throughout.
Yes, it is an AP.
Q1(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.
If initial air = V, then successive amounts are V, 3V/4, 9V/16, … The difference between consecutive terms is not constant (it keeps shrinking proportionally).
No, not an AP (it is a GP).
Q2(i). Write the first four terms of the AP when a = 10, d = 10.
10, 20, 30, 40.
Q2(iii). Write the first four terms of the AP when a = 4, d = −3.
4, 1, −2, −5.
Q3(i). For the AP 3, 1, −1, −3, …, write the first term and common difference.
a = 3, d = 1 − 3 = −2.
Q4. Which of the following are APs? If they form an AP, find the common difference and write three more terms. (i) 2, 4, 8, 16, … (iv) −10, −6, −2, 2, … (viii) −1/2, −1/2, −1/2, … (xii) √2, √8, √18, √32, …
(i) 4−2=2, 8−4=4 — differences not equal. Not an AP.
(iv) −6−(−10)=4, −2−(−6)=4, 2−(−2)=4. AP, d = 4; next terms: 6, 10, 14.
(viii) Every term is −1/2, so the difference between consecutive terms is 0. AP, d = 0; next terms: −1/2, −1/2, −1/2.
(xii) √8 = 2√2, √18 = 3√2, √32 = 4√2 — so the sequence is √2, 2√2, 3√2, 4√2, each differing by √2. AP, d = √2; next terms: 5√2, 6√2, 7√2.
Exercise 5.2
Q2(ii). Choose the correct choice: 11th term of the AP: −3, −1/2, 2, … is (A) 28 (B) 22 (C) −38 (D) −48 ½
a = −3, d = −1/2 − (−3) = 5/2. aₓ = a + (n−1)d, so a₁₁ = −3 + 10(5/2) = −3 + 25 = 22.
Answer: (B) 22.
Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a + 10d = 38 and a + 15d = 73. Subtracting: 5d = 35, so d = 7, and a = 38 − 70 = −32.
a₃₁ = a + 30d = −32 + 210 = 178.
Q11. Which term of the AP 3, 15, 27, 39, … will be 132 more than its 54th term?
d = 12. The nth and 54th terms differ by (n−54)d. Set (n−54)(12) = 132, so n−54 = 11, n = 65.
The 65th term.
Q17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Reverse the AP: last term becomes first term (a=253) with common difference −5 (since the original d=5). 20th term from the end = 253 + 19(−5) = 253 − 95 = 158.
Exercise 5.3
Q1(a). Find the sum: 2, 7, 12, …, to 10 terms.
a=2, d=5, n=10. Sₙ = (n/2)[2a+(n−1)d] = 5[4+45] = 5(49) = 245.
Q3(i). In an AP, given a=5, d=3, aₙ=50, find n and Sₙ.
aₙ = a+(n−1)d → 50 = 5+(n−1)3 → n−1=15 → n=16. S₁₆ = (16/2)(5+50) = 8(55) = 440.
Q5. Find the sum of the first 22 terms of an AP in which d = 7 and 22nd term is 149.
a₂₂ = a+21(7) = 149 → a = 2. S₂₂ = (22/2)(a+a₂₂) = 11(2+149) = 1661.
Q9. If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289, find the sum of the first n terms.
(7/2)[2a+6d]=49 → 2a+6d=14 → a+3d=7.
(17/2)[2a+16d]=289 → 2a+16d=34 → a+8d=17.
Subtracting: 5d=10 → d=2, a=1. Sₙ = (n/2)[2+(n−1)2] = (n/2)(2n) = n².
Q19. A ladder has rungs 25 cm apart, decreasing uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 ½ m apart, what is the length of the wood required for the rungs?
Number of gaps = 250/25 = 10, so there are 11 rungs. Lengths form an AP from 45 to 25: a=45, a₁₁=25, n=11. Sum = n(a+l)/2 = 11(45+25)/2 = 11(35) = 385 cm.
Related: Solutions | Extra Questions | Revision Notes | Formulas Handbook | Class 10 Maths Book
