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UPSC NDA II 2018 Mathematics Solved Paper

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Question : 7 of 120

Marks: +1, -0

The sum of the series

3-1+\frac{1}{3}-\frac{1}{9}+…

is equal to

$20/9$
$9/20$
$9/4$
$4/9$

Solution:

Concept:

The sum of an infinite geometric progression (GP) is given by

S_{\infty} = \frac{a}{1 - r}

, provided that

|r| < 1

.

Explanation:

The given series is

3 - 1 + \frac{1}{3} - \frac{1}{9} + \dots

.

It is a GP with first term

a = 3

and common ratio

r = \frac{-1}{3}

.

Since

|r| = \frac{1}{3} < 1

, the infinite sum formula applies:

S_{\infty} = \frac{3}{1 - \left(-\frac{1}{3}\right)} = \frac{3}{1 + \frac{1}{3}} = \frac{3}{\frac{4}{3}} = \frac{9}{4}

.

Answer:

\frac{9}{4}

(Option C).

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