If 8 = 3+1/4(3+p) + 1/4^2(3 + 2p) + 1/4^3(3 + 3p) + …∞, then the value of p is ____.

Question

If \(8=3+\frac{1}{4}(3+p)+\frac{1}{4^{2}}(3+2 p)+\frac{1}{4^{3}}(3+3 p)+\ldots \infty\), then the value of p is _____.

Answer 1

Correct answer: 9

\(8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^{2}}\)

\(\left(\text{sum of infinite terms of A.G.P} =\frac{a}{1-r}+\frac{d r}{(1-r)^{2}}\right)\)

\(\Rightarrow \frac{4 \mathrm{p}}{9}=4 \Rightarrow \mathrm{p}=9\)