Sequence and SeriesMCQMTP June 24 Series IIQuestion 1819 of 213
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Divide 144 into three parts which are in AP and such that the largest is twice the smallest, the smallest of three numbers will be:

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option d — 32

All Options:

  • A48
  • B32
  • C13
  • D32

Detailed Solution & Explanation

Let the three parts in A.P. be:
a−d, a, a+da-d, \ a, \ a+d
Since their sum is $144$:
(a−d)+a+(a+d)=144(a-d) + a + (a+d) = 144
3a=144  ⟹  a=483a = 144 \implies a = 48

We are given that the largest part is twice the smallest part:
a+d=2(a−d)a+d = 2(a-d)
a+d=2a−2da+d = 2a - 2d
3d=a  ⟹  3d=48  ⟹  d=163d = a \implies 3d = 48 \implies d = 16

The smallest number is:
a−d=48−16=32a-d = 48 - 16 = 32
Hence, **Option D** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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