The sum of infinity of the series

Question:

The sum of infinity of the series $9-3+1-\frac{1}{3}+$ , is

Solution:

$9-3+1-\frac{1}{3}+\ldots \ldots \ldots \ldots$

Here $a=9$

$r=\frac{a_{2}}{a_{1}}=\frac{-3}{9}=\frac{-1}{3}$

$r=\frac{a_{3}}{a_{2}}=\frac{1}{-3}=\frac{-1}{3}$

Sum of infinite terms of G.P is

$S_{n}=\frac{a}{1-r}$

$S_{n}=\frac{9}{1-\left(\frac{-1}{3}\right)}$

$=\frac{9}{\frac{3+1}{3}}$

$=\frac{9 \times 3}{4}$

$S_{n}=\frac{27}{4}$

i.e the sum of infinity of the series

 

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