JEE Main 2023 — Sequences And Series Question with Solution

Given the conditions :

1. $a_6+a_8=2\Rightarrow a{r}^5+a{r}^7=2$
2. $a_3\cdot a_5=\frac{1}{9}\Rightarrow a^2\cdot r^2\cdot r^4=\frac{1}{9}\Rightarrow a{r}^3=\frac{1}{3}$

From this, we can form the equation $\frac{r^2}{3}+\frac{r^4}{3}=2$, which simplifies to $r^4+r^2=6$.

This can be factored to give $\left(r^2-2\right)\left(r^2+3\right)=0$, yielding $r^2=2$ (since $r^2$ cannot be $-3$ for real $r$).

So, we have $r=\sqrt{2}$.

Substituting $r=\sqrt{2}$ into the equation $ar=\frac{1}{6}$, we get $a=\frac{1}{6\sqrt{2}}$.

Now, we find the value of $6(a_2+a_4)(a_4+a_6)$:

$6(a_2+a_4)(a_4+a_6)=6\left(ar+a{r}^3\right)\left(a{r}^3+a{r}^5\right)$

$=6\left(\frac{1}{6\sqrt{2}}+\frac{1}{3\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}}+\frac{2}{3\sqrt{2}}\right)$

$=6\cdot \frac{1}{2}\cdot 1=3$.