Let a _ 1 , a _ 2 , a _ 3 , . be a G.P. of increasing positive...

Given,

${a}_{1}, {a}_{2}, {a}_{3}, \ldots$. be a G.P. of increasing positive numbers,

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

So, ${a}_{6} + {a}_{8} = 2$

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

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

So, ${a}_{3} \cdot {a}_{5} = \dfrac{1}{9} ⇒ {a}^{2} \cdot {r}^{2} \cdot {r}^{4} = \dfrac{1}{9}$

$⇒ a{r}^{3} = \dfrac{1}{3}$

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

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

$⇒ {r}^{4} + {r}^{2} = 6$

$⇒ \left({r}^{2} + 3\right)\left({r}^{2} - 2\right) = 0$

$⇒ {r}^{2} = 2$

$a{r}^{3} = \dfrac{1}{3} ⇒ ar \cdot 2 = \dfrac{1}{3} ⇒ ar = \dfrac{1}{6}$

Now finding the value of $6\left({a}_{2} + {a}_{4}\right)\left({a}_{4} + {a}_{6}\right)$, we get

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

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

$= 6 \cdot \dfrac{1}{2} \cdot 1 = 3$