If the sum of an infinite GP a,

Question:

If the sum of an infinite GP a, ar, ar $^{2}, a r^{3}, \ldots$ is 15 and the sum of the squares of its each term is 150 , then the sum of $\operatorname{ar}^{2}, \operatorname{ar}^{4}, \operatorname{ar}^{6}, \ldots$ is :

  1. $\frac{5}{2}$

  2. $\frac{1}{2}$

  3. $\frac{25}{2}$

  4. $\frac{9}{2}$


Correct Option: , 2

Solution:

Sum of infinite terms :

$\frac{a}{1-r}=15$ ............(i)

Series formed by square of terms:

$a^{2}, a^{2} r^{2}, a^{2} r^{4}, a^{2} r^{6} \quad \ldots .$

$\operatorname{Sum}=\frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=150$

$\Rightarrow \frac{\mathrm{a}}{1-\mathrm{r}} \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150 \Rightarrow 15 \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150$

$\Rightarrow \frac{\mathrm{a}}{1+\mathrm{r}}=10$ ...........(ii)

by (i) and (ii) $a=12 ; r=\frac{1}{5}$

Now series : $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}$

$\operatorname{Sum}=\frac{\mathrm{ar}^{2}}{1-\mathrm{r}^{2}}=\frac{12 \cdot(1 / 25)}{1-1 / 25}=\frac{1}{2}$

 

 

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