Sequence and SeriesPYQ May 25Question 4331 of 150
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Find the sum of series 1+12+14+…\displaystyle 1 + \frac{1}{2} + \frac{1}{4} + \dots upto 6 terms.

Options

A6332\displaystyle \frac{63}{32}
B3263\displaystyle \frac{32}{63}
C2653\displaystyle \frac{26}{53}
D5326\displaystyle \frac{53}{26}
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Correct Answer

✅ Option a — 6332\displaystyle \frac{63}{32}

All Options:

  • A6332\displaystyle \frac{63}{32}
  • B3263\displaystyle \frac{32}{63}
  • C2653\displaystyle \frac{26}{53}
  • D5326\displaystyle \frac{53}{26}

Detailed Solution & Explanation

The given series is 1+12+14+…\displaystyle 1 + \frac{1}{2} + \frac{1}{4} + \dots
This is a Geometric Progression (G.P.) with:
- First term (a\displaystyle a) = 1\displaystyle 1
- Common ratio (r\displaystyle r) = 1/21=12\displaystyle \frac{1/2}{1} = \frac{1}{2}
- Number of terms (n\displaystyle n) = 6\displaystyle 6

Since the common ratio is less than 1\displaystyle 1 (r<1\displaystyle r < 1), we use the sum formula:
Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r}
Substitute a=1,r=12,n=6\displaystyle a = 1, r = \frac{1}{2}, n = 6 into the formula:
S6=1×(1−(12)6)1−12S_6 = \frac{1 \times \left(1 - \left(\frac{1}{2}\right)^6\right)}{1 - \frac{1}{2}}
S6=1−16412S_6 = \frac{1 - \frac{1}{64}}{\frac{1}{2}}
S6=2×(6364)=6332S_6 = 2 \times \left(\frac{63}{64}\right) = \frac{63}{32}
Hence, **Option A** 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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