Mathematics - Sequence & Series Question with Solution | TestHub

Let $S=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\dots$. This is an Arithmetico-Geometric Progression. The arithmetic progression part is $1,4,7,10,\dots$ with common difference $3$, and the geometric progression part is $1,\frac{1}{5},\frac{1}{5^2},\frac{1}{5^3},\dots$ with common ratio $r=\frac{1}{5}$. Multiply $S$ by $r=\frac{1}{5}$: $$\frac{1}{5}S=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\dots$$ Subtract the second equation from the first: $$\left(1-\frac{1}{5}\right)S=1+\left(\frac{4}{5}-\frac{1}{5}\right)+\left(\frac{7}{5^2}-\frac{4}{5^2}\right)+\left(\frac{10}{5^3}-\frac{7}{5^3}\right)+\dots$$ $$\frac{4}{5}S=1+\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+\dots$$

The terms from $\frac{3}{5}$ onwards form an infinite G.P. with first term $A=\frac{3}{5}$ and common ratio $R=\frac{1}{5}$. Its sum is $\frac{A}{1-R}=\frac{3/5}{1-1/5}=\frac{3/5}{4/5}=\frac{3}{4}$. So, $\frac{4}{5}S=1+\frac{3}{4}=\frac{7}{4}$. Therefore, $S=\frac{7}{4}\times \frac{5}{4}=\frac{35}{16}$.