Let a₁, a₂, a₃, …….. be a G.P. of increasing positive numbers. Let the sum of its 6^th and 8^th terms be 2 the product of its 3^rd and 5^th terms be 1/9. Then 6(a₂ + a₄)(a₄ + a₆) a is equal to

$\begin{array}{l}6\left({\mathrm{a}}_2+{\mathrm{a}}_4\right)\left({\mathrm{a}}_4+{\mathrm{a}}_6\right)=3\\ {\mathrm{T}}_6+{\mathrm{T}}_8={\mathrm{a}}_1{\mathrm{r}}^5+{\mathrm{a}}_1{\mathrm{r}}^7=2\\ \Rightarrow {\mathrm{a}}_1{\mathrm{r}}^5\left(1+{\mathrm{r}}^2\right)=2 ..........\left(1\right)\\ {\mathrm{T}}_3\cdot {\mathrm{T}}_5=\frac{1}{9}\Rightarrow {\mathrm{a}}_1{\mathrm{r}}^2\cdot {\mathrm{a}}_1{\mathrm{r}}^4=\frac{1}{9}\Rightarrow {\mathrm{a}}_1^2{\mathrm{r}}^6=\frac{1}{9}\\ \Rightarrow {\mathrm{a}}_1{\mathrm{r}}^3=\frac{1}{3} ......\left(2\right)\\ \Rightarrow {\mathrm{a}}_1{\mathrm{r}}^3{\mathrm{r}}^2\left(1+{\mathrm{r}}^2\right)=2\Rightarrow \frac{1}{3}{\mathrm{r}}^2\left(1+{\mathrm{r}}^2\right)=2\\ \Rightarrow {\mathrm{r}}^4+{\mathrm{r}}^2-6=0\Rightarrow \left({\mathrm{r}}^2+3\right)\left({\mathrm{r}}^2-2\right)=0\Rightarrow {\mathrm{r}}^2=2\\ {\mathrm{a}}_1=\frac{1}{6\sqrt{2}}\\ 6\left({\mathrm{a}}_2+{\mathrm{a}}_4\right)\left({\mathrm{a}}_4+{\mathrm{a}}_6\right)=6\left({\mathrm{a}}_1\mathrm{r}+{\mathrm{a}}_1{\mathrm{r}}^3\right)\left({\mathrm{a}}_1{\mathrm{r}}^3+{\mathrm{a}}_1{\mathrm{r}}^5\right)=6{\mathrm{a}}_1^2{\mathrm{r}}^4{\left(1+{\mathrm{r}}^2\right)}^2=3\end{array}$