Explanation
\begin{aligned}
& 3+\frac{1}{4}(3+p)+\frac{1}{4^{2}}(3+2p)+\frac{1}{4^{3}}(3+3o)+...\infty=8 \\
& \Rightarrow\left(3+\frac{3}{4}+\frac{3}{4^{2}}+...\right)+\left(\frac{p}{4}+\frac{2p}{4^{2}}+\frac{3p}{4^{3}}+....\right)=8 \\
& \Rightarrow\left[\frac{3}{1-\frac{1}{4}}\right]+\left[\frac{p}{4}+\frac{2p}{4^{2}}+\frac{3p}{4^{3}}+..\right]=8 \\
& \Rightarrow\frac{p}{4}+\frac{2p}{4^{2}}+\frac{3p}{4^{3}}+...=4 \\
& \mathrm{Let}x=\frac{p}{4}+\frac{2p}{4^{2}}+\frac{3p}{4^{3}}+... \\
& Ondividingby4 \\
& \frac{x}{4}=\frac{p}{4^{2}}+\frac{2p}{4^{3}}+\frac{3p}{4^{4}}+... \\
& \therefore\left(x-\frac{x}{4}\right)=\left(\frac{p}{4}\right)+\left(\frac{2p}{4^{2}}-\frac{p}{4^{2}}\right)+\left(\frac{3p}{4^{3}}-\frac{2p}{4^{3}}\right)+... \\
& \Rightarrow\frac{3x}{4}=\frac{p}{4}+\frac{p}{4^{2}}+\frac{p}{4^{3}}+....=\left[\frac{\frac{p}{4}}{1-\frac{1}{4}}\right]=\frac{p}{3} \\
& \Rightarrow\frac{p}{9}=4\Rightarrow p=q
\end{aligned}
