If \(8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}\left(3+p^2\right)+\ldots \infty\) then the value of \(p\) is (22 Jan, Shift I, M…

\(\begin{aligned}
& 8=\left(3+\frac{3}{4}+\frac{3}{4^2}+\ldots+\infty\right)+\left(\frac{p}{4}+\frac{p^2}{4^2}+\ldots+\infty\right) \\
& 8=3\left(1+\frac{1}{4}+\frac{1}{4^2}+\ldots+\infty\right)+\left(\frac{p}{4}+\frac{p^2}{4^2}+\ldots+\infty\right) \\
& 8=3\left(\frac{1}{1-\frac{1}{4}}\right)+\frac{\frac{p}{4}}{1-\frac{p}{4}}
\end{aligned}\)

\(\begin{aligned}
& 8=3\left(\frac{4}{3}\right)+\frac{p}{4-p} \\
& 4=\frac{p}{4-p} \\
& \Rightarrow 16-4 p=p \\
& \Rightarrow 5 p=16 \\
& \Rightarrow p=\frac{16}{5}
\end{aligned}\)