Sequence and SeriesPYQ Sept 25Question 4131 of 150
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The sum of first two terms of a geometric progression is 14 and its infinite sum i.e. sum up to infinity is 32. What is the common ratio of the progression?

Options

A0.5
B0.75
C1.25
D0.25
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Correct Answer

✅ Option b — 0.75

All Options:

  • A0.5
  • B0.75
  • C1.25
  • D0.25

Detailed Solution & Explanation

Let the first term of the geometric progression be a\displaystyle a and the common ratio be r\displaystyle r (where ∣r∣<1\displaystyle |r| < 1).
1) The sum of the first two terms is 14: a+ar=14  ⟹  a(1+r)=14a + ar = 14 \implies a(1+r) = 14
2) The sum up to infinity is 32: S∞=a1−r=32  ⟹  a=32(1−r)S_{\infty} = \frac{a}{1-r} = 32 \implies a = 32(1-r)
Substitute the expression for a\displaystyle a from the second equation into the first: 32(1−r)(1+r)=1432(1-r)(1+r) = 14 32(1−r2)=1432(1-r^2) = 14 1−r2=1432=7161-r^2 = \frac{14}{32} = \frac{7}{16} r2=1−716=916r^2 = 1 - \frac{7}{16} = \frac{9}{16} Taking the square root on both sides: r=±34=±0.75r = \pm \frac{3}{4} = \pm 0.75
Among the options, the positive common ratio is 0.75\displaystyle 0.75.
Hence, **Option B** 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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