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UPSC NDA Math Model Paper 1

Question : 2 of 120

Marks: +1, -0

\; \text{ The value of } \; 9 \; \frac{1}{3} 9 \; \frac{1}{9} 9 \; \frac{1}{27} \dots \infty \; \text{ is: } \;

Solution:

Concept:

Geometric Progression (GP):

The series of numbers where the ratio of any two consecutive terms is the same is called a Geometric Progression.
A Geometric Progression of n terms with first term a and common ratio $r$ is represented as: a, ar, ar $^{2}, a r^{3}, \dots, a r^{n-2}, a r^{n-1}.$
The sum of the first $n$ terms of a GP is: $S_{n}=a\left( \frac{r^{n}-1}{r-1} \right)$ .
The sum to $\infty$ of a GP, when $|r|<1,$ is: $S_{\infty}=\frac{a}{1-r}$ .

Calculation:

Let us consider the infinite series

\; \frac{1}{3} + \; \frac{1}{9} + \; \frac{1}{27} + \dots \infty

Here,

a=\frac{1}{3}

and

r = \frac{ \frac{1}{9} }{ \frac{1}{3} } = \frac{1}{3}

\therefore S_{\infty}= \frac{a}{1-r} = \frac{ \frac{1}{3} }{ 1 - \frac{1}{3} } = \frac{ \frac{1}{3} }{ \frac{2}{3} } = \frac{1}{2}

Now, let

P = 9^{ \frac{1}{3} } 9^{ \frac{1}{9} } 9 \; \frac{1}{27} \dots \infty

\therefore P = 9^{ \frac{1}{3} } + \frac{1}{9} + \frac{1}{27} + \dots \infty = 9^{ \frac{1}{2} } = \sqrt{9} = 3 .

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