CBSE • Class 10 • Mathematics • Chapter 5 (Arithmetic Progressions) • Exercise 5.3

Exercise 5.3: Arithmetic Progressions — NCERT Solutions

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

Aligned to the latest NCERT 2024-25 edition • 20 questions in this exercise • Free plan, no credit card

Open this exercise in the AI tutorBack to Chapter 5

What this exercise covers

Sum formulaSum of first n natural numbersWord problems on instalments

Step-by-step solutions — Exercise 5.3

3 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 5.3 Q1 • 3 marks

Find the sum of the first 20 terms of the AP 5,9,13,17,…5, 9, 13, 17, \dots
Hint (Socratic — try this first)▾
Which sum formula uses aa, dd and nn directly?
Step-by-step solution▾

Here a=5a = 5, d=4d = 4, n=20n = 20.

Using Sn=n2[2a+(n−1)d]S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]: S20=202[2(5)+(20−1)4]S_{20} = \frac{20}{2}\big[2(5) + (20-1)4\big] =10[10+76]=10×86=860= 10\big[10 + 76\big] = 10 \times 86 = 860

The sum is 860860.

Common mistake:
Multiplying n/2n/2 by only one part of the bracket, or using (n)(n) instead of (n−1)(n-1).
Open this question in the AI tutor →

Exercise 5.3 Q2 • 3 marks

Find the sum of the first 15 multiples of 8.
Hint (Socratic — try this first)▾
List the multiples; what are aa, dd and the number of terms?
Step-by-step solution▾

The first 15 multiples of 8 are 8,16,24,…8, 16, 24, \dots up to 8×15=1208 \times 15 = 120.

So a=8a = 8, d=8d = 8, n=15n = 15, last term l=120l = 120.

Using Sn=n2(a+l)S_n = \dfrac{n}{2}(a + l): S15=152(8+120)=152×128=15×64=960S_{15} = \frac{15}{2}(8 + 120) = \frac{15}{2}\times 128 = 15 \times 64 = 960

The sum is 960960.

Common mistake:
Confusing 'first 15 multiples' with 'multiples up to 15' and using the wrong last term.
Open this question in the AI tutor →

Exercise 5.3 Q3 • 4 marks

How many terms of the AP 9,17,25,…9, 17, 25, \dots must be taken so that their sum is 636636?
Hint (Socratic — try this first)▾
Set the sum formula equal to 636 and solve the resulting quadratic in nn.
Step-by-step solution▾

Here a=9a = 9, d=8d = 8, Sn=636S_n = 636.

Sn=n2[2a+(n−1)d]=636S_n = \frac{n}{2}\big[2a + (n-1)d\big] = 636 n2[18+(n−1)8]=636\frac{n}{2}\big[18 + (n-1)8\big] = 636 n2[8n+10]=636\frac{n}{2}\big[8n + 10\big] = 636 n(4n+5)=636n(4n + 5) = 636 4n2+5n−636=04n^2 + 5n - 636 = 0

Solving: n=−5±25+101768=−5±1018n = \dfrac{-5 \pm \sqrt{25 + 10176}}{8} = \dfrac{-5 \pm 101}{8}.

Taking the positive value: n=968=12n = \dfrac{96}{8} = 12.

So 12 terms are required.

Common mistake:
Accepting the negative root of the quadratic as a valid number of terms.
Open this question in the AI tutor →

How to approach Exercise 5.3

  1. Re-read the chapter summary first. Open Arithmetic Progressions and refresh the key concepts: First term, Common difference, nth term, Sum of n terms.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Arithmetic Progressions

  1. Exercise 5.1 — Identifying arithmetic progressions and writing the common difference.
  2. Exercise 5.2 — nth term of an AP — applying aₙ = a + (n − 1)d in direct and word problems.
  3. Exercise 5.3 — Sum of first n terms — Sₙ = n/2 [2a + (n − 1)d].
  4. Exercise 5.4 — Mixed AP problems combining nth term and sum, including pattern problems.

← Exercise 5.2

nth term of an AP — applying aₙ = a + (n − 1)d in direct and word problems.

Exercise 5.4 →

Mixed AP problems combining nth term and sum, including pattern problems.

Solve Exercise 5.3 with AI guidance

Free plan. No credit card. Works on any device.

Start Free