In an increasing geometric series, the sum of the second and the sixth term is 25 2 and the product of the…

$T_2+T_6=\frac{25}{2}\Rightarrow \mathit{ar}\left(1+r^4\right)=\frac{25}{2}$

$a^2{r}^2{\left(1+r^4\right)}^2=\frac{625}{4} \cdots \left(1\right)$

$T_3·T_5=25\Rightarrow \left(a{r}^2\right)\left(a{r}^4\right)=25$

$a^2{r}^6=25 ...\left(2\right)$

On dividing $\left(1\right)$ by $\left(2\right)$

$\frac{{\left(1+r^4\right)}^2}{r^4}=\frac{25}{4}$

$4r^8-17r^4+4=0$

$\left(4r^4-1\right)\left(r^4-4\right)=0$

$r^4=\frac{1}{4},4\Rightarrow r^4=4$ (an increasing geometric series)

$a^2{r}^6=25\Rightarrow {\left(a{r}^3\right)}^2=25$

$T_4+T_6+T_8=a{r}^3+a{r}^5+a{r}^7$

$={\mathit{ar}}^3\left(1+r^2+r^4\right)$

$=5\left(1+2+4\right)=35$