Question

Find sum of infinte AGP :
$\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+....\infty$

Answer 1

Given, $S=\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+....\left(1\right)$

Multiplying $S$ with $\frac{1}{3}$, we get

$\frac{S}{3}=\frac{3}{15}+\frac{5}{45}+\frac{7}{135}+....\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$, we get

$S=\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+....\infty -\underset{–}{\frac{S}{3}=\frac{3}{15}+\frac{5}{45}+\frac{7}{135}+......\infty }\frac{2}{3}S=\frac{3}{5}+\frac{2}{15}+\frac{2}{45}+\frac{2}{135}+....\infty \frac{2}{3}S=\frac{3}{5}+2[\frac{1}{15}+\frac{1}{45}+\frac{1}{135}+....\infty ]$

It is an Infinite GP with $r=\frac{1}{3}$

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

$\frac{2}{3}S=\frac{3}{5}+2\frac{1/15}{(1-\frac{1}{3})}\frac{2}{3}S=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}S=\frac{6}{5}$