In a increasing geometric series,

Solution:

$\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \ldots$

$\mathrm{T}_{2}+\mathrm{T}_{6}=\frac{25}{2} \Rightarrow \operatorname{ar}\left(1+\mathrm{r}^{4}\right)=\frac{25}{2}$

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

$\mathrm{T}_{3} \cdot \mathrm{T}_{5}=25 \Rightarrow\left(\mathrm{ar}^{2}\right)\left(a r^{4}\right)=25$

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

On dividing (1) by (2)

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

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

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

$r^{4}=\frac{1}{4}, 4 \Rightarrow r^{4}=4$

(an increasing geometric series)

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

$\mathrm{T}_{4}+\mathrm{T}_{6}+\mathrm{T}_{8}=\mathrm{ar}^{3}+\mathrm{ar}^{5}+\mathrm{ar}^{7}$

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

$=5(1+2+4)=35$