Sequence and SeriesMCQPYQ Dec. 21Question 1833 of 212
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The largest value of n\displaystyle n for which 1+12+122++12n1<0.998\displaystyle 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}} < 0.998 is

Options

A9
B6
C7
D8
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Correct Answer

Option d8

All Options:

  • A9
  • B6
  • C7
  • D8

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Detailed Solution & Explanation

Let us compute the sum of the geometric progression 12+14++12n\displaystyle \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n}:
Sn=1(12)nS_n = 1 - \left(\frac{1}{2}\right)^n
We want to find the largest n\displaystyle n such that:
1(12)n<0.9981 - \left(\frac{1}{2}\right)^n < 0.998
(12)n>0.002=1500\left(\frac{1}{2}\right)^n > 0.002 = \frac{1}{500}
2n<5002^n < 500

We check powers of 2:
28=256<5002^8 = 256 < 500
29=512>5002^9 = 512 > 500
So the largest value of n\displaystyle n is 8\displaystyle 8.
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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