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Sum of n Terms of a G.P. - Class XI

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

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$\dfrac{3}{2}$

$3$

💡 Explanation:

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

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

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

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

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

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

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

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