QJEEM23Mad8ea: JEE Main 2023 Mathematics - Class 11 JEEM23Mad8ea Solution: Let a 1 , a 2 , a 3 , .... be a G.P. of increasing positive…

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 $a r = \frac{1}{6}$ , we get $a = \frac{1}{6\sqrt{2}}$ .

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

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

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

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