Extra Questions: Class 10 Maths Chapter 5 Arithmetic Progressions

Fresh, reasoning-heavy AP problems beyond the standard exercise. See the standard-level Solutions first if you haven’t already. These Class 10 Mathematics Chapter 5 important questions are handy for last-minute exam practice.

Last Updated: October 24, 2026

How to Approach the HOTS Questions in This Chapter

Extra Questions (HOTS Level): Arithmetic Progressions (Class 10 Maths Chapter 5)

Q1. The sum of three consecutive terms of an AP is 36, and the sum of their squares is 464. Find the terms.
Let the terms be (a−d), a, (a+d). Sum = 3a = 36, so a = 12.
Sum of squares: (a−d)²+a²+(a+d)²=464 → 3a²+2d²=464 → 3(144)+2d²=464 → 2d²=32 → d²=16 → d=4.
Terms: 8, 12, 16.

Q2. Assertion-Reason: Assertion (A): The sequence 1, 4, 9, 16, 25, … is an AP. Reason (R): A sequence is an AP if the difference between any two consecutive terms is constant. Choose: (i) Both A and R true, R explains A (ii) Both true, R doesn’t explain A (iii) A true, R false (iv) A false, R true.
The given sequence is squares of natural numbers (1²,2²,3²,4²,5²); differences are 3, 5, 7, 9 — not constant, so it is NOT an AP. R is a true, correct definition of an AP.
Answer: (iv) A is false, R is true.

Q3. A construction company stacks pipes so that each row has one fewer pipe than the row below, starting with 20 pipes in the bottom row and ending with 1 pipe at the top. How many pipes are there in total, and how many rows?
This is an AP: 20, 19, 18, …, 1 with a=20, d=−1, l=1. Number of terms: 1=20+(n−1)(−1) → n=20. Sum = n(a+l)/2 = 20(21)/2 = 210 pipes in 20 rows.

Q4. If the mth term of an AP is 1/n and the nth term is 1/m, show that the sum of the first mn terms is (mn+1)/2.
a+(m−1)d = 1/n and a+(n−1)d = 1/m. Subtracting: (m−n)d = 1/n − 1/m = (m−n)/mn, so d = 1/mn.
Substituting back: a = 1/n − (m−1)/mn = (m − (m−1))/mn = 1/mn.
S_{mn} = (mn/2)[2a+(mn−1)d] = (mn/2)[2/mn + (mn−1)/mn] = (mn/2)·(mn+1)/mn = (mn+1)/2. (Shown.)

Q5. A hall has 30 rows. The first row has 20 seats, and each subsequent row has 2 more seats than the previous row. Find the total seating capacity of the hall.
a=20, d=2, n=30. S₃₀ = (30/2)[2(20)+29(2)] = 15[40+58] = 15(98) = 1470 seats.

Related: Solutions | Extra Questions | Revision Notes | Formulas Handbook | Class 10 Maths Book

Frequently Asked Questions

How can you determine whether a given sequence of numbers is an arithmetic progression?
A sequence is an arithmetic progression if the difference between every pair of consecutive terms remains exactly the same throughout, which can be verified by subtracting each term from the one that follows it.

What real-life pattern of increase or decrease can be modelled using an arithmetic progression?
Situations involving a constant, fixed increase or decrease at regular intervals, such as a salary that increases by a fixed amount every year, can be modelled using an arithmetic progression.

Written by Satish

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