Mathematics - Sequence & Series Question with Solution | TestHub

$S=a+(a+d)r+(a+2d)r^2+\dots$ Then $S=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2},|r|<1$ Since, $r=\frac{1}{7}$ and $a=5,d=\alpha$ $\begin{array}{rl}& 7=\frac{5}{1-\frac{1}{7}}+\frac{\alpha \cdot \frac{1}{7}}{{\left(1-\frac{1}{7}\right)}^2}\\ & \Rightarrow \alpha =6\end{array}$