Question

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

Answer 1

Correct Answer: 9

Given series:
\(8=3+\dfrac{1}{4}(3+p)+\dfrac{1}{4^2}(3+2p)+\dfrac{1}{4^3}(3+3p)+\ldots\)
This is an arithmetic-geometric progression (A.G.P). Using the sum formula for an infinite
A. G. P. ,  we have:
\(\text{Sum}=\frac{a}{1-r}+\frac{d\cdot r}{(1-r)^2}\)
Solving for \(p:\)
\(\dfrac{4p}{9}=4\Rightarrow p=9\)