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

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 $p:$
$\dfrac{4p}{9}=4\Rightarrow p=9$