Let $a_1,a_2,a_3,\dots$. 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 $6\left(a_2+a_4\right)\left(a_4+a_6\right)$ is equal to

Given,

$a_1,a_2,a_3,\dots$. be a G.P. of increasing positive numbers,

And the sum of its $6^{\text{th }}$ and $8^{\text{th }}$ terms be $2$

So, $a_6+a_8=2$

$\Rightarrow a{r}^5+a{r}^7=2 .....\left(1\right)$

And the product of its $3^{\text{rd }}$ and $5^{\text{th }}$ terms be $\frac{1}{9}$,

So, $a_3·a_5=\frac{1}{9}\Rightarrow a^2·r^2·r^4=\frac{1}{9}$

$\Rightarrow a{r}^3=\frac{1}{3}$

Now from putting the value of $a{r}^3=\frac{1}{3}$ in equation $\left(1\right)$ we get,

$\frac{r^2}{3}+\frac{r^4}{3}=2$

$\Rightarrow r^4+r^2=6$

$\Rightarrow \left(r^2+3\right)\left(r^2-2\right)=0$

$\Rightarrow r^2=2$

$a{r}^3=\frac{1}{3}\Rightarrow ar·2=\frac{1}{3}\Rightarrow ar=\frac{1}{6}$

Now finding the value of $6\left(a_2+a_4\right)\left(a_4+a_6\right)$, we get

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

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

$=6·\frac{1}{2}·1=3$