CBSE • Class 10 • Mathematics • Chapter 5 (Arithmetic Progressions) • Exercise 5.4

Exercise 5.4: Arithmetic Progressions — NCERT Solutions

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

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What this exercise covers

Mixed AP word problemsPattern problemsReal-world applications

Step-by-step solutions — Exercise 5.4

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

Exercise 5.4 Q1 • 4 marks

In an AP, the sum of the first nn terms is given by Sn=3n2+5nS_n = 3n^2 + 5n. Find its nnth term and hence its 10th term.
Hint (Socratic — try this first)▾
How is the nnth term related to SnS_n and Sn−1S_{n-1}?
Step-by-step solution▾

The nnth term is an=Sn−Sn−1a_n = S_n - S_{n-1}.

Sn=3n2+5nS_n = 3n^2 + 5n Sn−1=3(n−1)2+5(n−1)=3(n2−2n+1)+5n−5=3n2−6n+3+5n−5S_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 Sn−1=3n2−n−2S_{n-1} = 3n^2 - n - 2

Therefore: an=(3n2+5n)−(3n2−n−2)=6n+2a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2

The 10th term: a10=6(10)+2=62a_{10} = 6(10) + 2 = 62

Common mistake:
Errors in expanding Sn−1S_{n-1}, especially mishandling the (n−1)2(n-1)^2 term or sign of the subtraction.
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Exercise 5.4 Q2 • 4 marks

A spiral is made of successive semicircles with radii 0.5,1.0,1.5,2.0,…0.5, 1.0, 1.5, 2.0, \dots cm. Find the total length of the first 13 semicircles. (Take π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
The length of a semicircle is πr\pi r; do the lengths themselves form an AP?
Step-by-step solution▾

Length of a semicircle of radius rr is πr\pi r.

Lengths: 0.5π,1.0π,1.5π,…0.5\pi, 1.0\pi, 1.5\pi, \dots — an AP with a=0.5πa = 0.5\pi, d=0.5πd = 0.5\pi, n=13n = 13.

Sum: S13=132[2(0.5π)+(13−1)(0.5π)]S_{13} = \frac{13}{2}\big[2(0.5\pi) + (13-1)(0.5\pi)\big] =132[π+6π]=132×7π=91π2= \frac{13}{2}\big[\pi + 6\pi\big] = \frac{13}{2}\times 7\pi = \frac{91\pi}{2}

With π=227\pi = \tfrac{22}{7}: S13=912×227=91×2214=200214=143 cmS_{13} = \frac{91}{2}\times \frac{22}{7} = \frac{91 \times 22}{14} = \frac{2002}{14} = 143 \text{ cm}

Total length is 143143 cm.

Common mistake:
Using the circumference 2πr2\pi r instead of the semicircle length πr\pi r, doubling the answer.
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How to approach Exercise 5.4

  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 5.3

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

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