Question

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

Answer 1

$$\begin{gathered} 9-3+1-\frac{1}{3}+.......... \\ \mathrm{Here}a=9 \\ r=\frac{a_2}{a_1}=\frac{-3}{9}=\frac{-1}{3} \\ r=\frac{a_3}{a_2}=\frac{1}{-3}=\frac{-1}{3} \end{gathered}$$
Sum of infinite terms of G.P is
$\begin{array}{rcl}{S}_n& =& \frac{a}{1-r}\\ S_n& =& \frac{9}{1-\left({\displaystyle \frac{-1}{3}}\right)}\\ & =& \frac{9}{{\displaystyle \frac{3+1}{3}}}\\ & =& \frac{9\times 3}{4}\end{array}$
$\begin{array}{rcl}& & S_n=\frac{27}{4}\end{array}$
i.e the sum of infinity of the series