Sequence and SeriesMCQMTP June 24 Series IIQuestion 1819 of 212
All Questions

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:

Options

A48
B32
C13
D32
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Correct Answer

Option d32

All Options:

  • A48
  • B32
  • C13
  • D32

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Detailed Solution & Explanation

Let the three parts in A.P. be:
ad, a, a+da-d, \ a, \ a+d
Since their sum is 144\displaystyle 144:
(ad)+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(ad)a+d = 2(a-d)
a+d=2a2da+d = 2a - 2d
3d=a    3d=48    d=163d = a \implies 3d = 48 \implies d = 16

The smallest number is:
ad=4816=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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