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8. Sequences and Series
hard
Let $a _1, a _2, a _3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text {th }}$ and $8^{\text {th }}$ terms be $2$ and the product of its $3^{\text {rd }}$ and $5^{\text {th }}$ terms be $\frac{1}{9}$. Then $6\left( a _2+\right.$ $\left.a_4\right)\left(a_4+a_6\right)$ is equal to
A
$2 \sqrt{2}$
B
$2$
C
$3 \sqrt{3}$
D
$3$
(JEE MAIN-2023)
Solution
$a r^5+a r^7=2$
$\left(a r^2\right)\left(a r^4\right)=\frac{1}{9}$
$a^2 r^6=\frac{1}{9}$
Now, $r > 0$
$\operatorname{ar}^5\left(1+r^2\right)=2$
Now, $ar ^3=\frac{1}{3}$ or $-\frac{1}{3}$ (rejected)
$r^2=2$
$r=\sqrt{2}$
$a=\frac{1}{6 \sqrt{2}}$
Now, $6\left(a_2+a_4\right)\left(a_4+a_6\right)$
$6\left(a r+a r^3\right)\left(a r^3+a r^5\right)$
$6 a^2 r^4\left(1+r^2\right)$
$6\left(\frac{1}{36.2}\right)(4)(9)=3$
Standard 11
Mathematics
