Question

If $1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\cdots \text{upto}\infty =\frac{a}{b}$ where $\text{H.C.F.}(a,b)=2$, then the value of $a-b$ is

Answer 1

$S_n=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\cdots \text{upto}\infty$
$\frac{S_n}{5}=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\frac{10}{5^4}+\cdots \text{upto}\infty$

$S_n-\frac{S_n}{5}=1+\frac{3}{5}+\frac{3}{5^2}+\cdots$
$\Rightarrow \frac{4S_n}{5}=1+\frac{\frac{3}{5}}{1-\frac{1}{5}}\Rightarrow S_n=\frac{35}{16}=\frac{a}{b}\because \text{H.C.F.}(a,b)=2,\Rightarrow (a,b)=(70,32)\Rightarrow a-b=38$