8. Sequences and Series
hard

एक अनंन्त $GPa , ar , a r ^{2}, a r ^{3}, \ldots$ का योग 15 है तथा इसके प्रत्येक पद का वर्ग करने का योग 150 है, तो $a r^{2}, a r^{4}, a r^{6}, \ldots$ का योग है।

A

$\frac{5}{2}$

B

$\frac{1}{2}$

C

$\frac{25}{2}$

D

$\frac{9}{2}$

(JEE MAIN-2021)

Solution

Sum of infinite terms :

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

Series formed by square of terms:

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

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

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

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

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

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

$\text { Sum }=\frac{a r^{2}}{1-r^{2}}=\frac{12 \cdot(1 / 25)}{1-1 / 25}=\frac{1}{2}$

Standard 11
Mathematics

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