Question

If the sum of infinite G.P. $p,1,\frac{1}{p},\frac{1}{p^2},.......,$ is $\frac{9}{2}$. Then find the value of $p$.

• A. $1$
• B. $\frac{3}{2}$
• C. $3$
• D. $\frac{5}{2}$

Answer 1

Answer: (B), (C)

$\frac{3}{2}$
$3$

Sum of infinite series of a GP $=\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio

Here $a=p$ and $r=\frac{1}{p}$

$\Rightarrow \frac{p}{1-\frac{1}{p}}=\frac{9}{2}$

$\Rightarrow \frac{p^2}{p-1}=\frac{9}{2}$

$\Rightarrow 2p^2-9p+9=0$

$\Rightarrow 2p^2-6p-3p+9=0$

$\Rightarrow (2p-3)(p-3)=0$

$\Rightarrow p=3,\frac{3}{2}$ (because common ratio $r=1/p$ must be less than 1)