If \(7=5+\frac{1}{7}(5+\alpha )+\frac{1}{{7}^{2}}(5+2\alpha )\)\(+\frac{1}{{7}^{3}}(5+3\alpha )+\ldots \ldots \ldots ...…

Let

\(S=5+\frac{1}{7}(5+\alpha)+\frac{1}{72}(5+2 \alpha)+\ldots \infty \quad \ldots(i)\)

On multiply by \(\frac{1}{7}\), we get

\(\frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^2}(5+\alpha)+\ldots \infty \ldots(i i)\)

Subtracting equation (ii) from (i), we get

\(\begin{aligned} & \frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right) \\ & 6=5+\frac{\alpha}{6} \\ & \Rightarrow \alpha=6\end{aligned}\)