In a increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25 . Then, the sum of $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is equal to
Correct Option: , 3
$\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \ldots$
$\mathrm{T}_{2}+\mathrm{T}_{6}=\frac{25}{2} \Rightarrow \operatorname{ar}\left(1+\mathrm{r}^{4}\right)=\frac{25}{2}$
$a^{2} r^{2}\left(1+r^{4}\right)^{2}=\frac{625}{4}$...(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)
$a^{2} r^{6}=25 \Rightarrow\left(a r^{3}\right)^{2}=25$
$\mathrm{T}_{4}+\mathrm{T}_{6}+\mathrm{T}_{8}=\mathrm{ar}^{3}+\mathrm{ar}^{5}+\mathrm{ar}^{7}$
$=a r^{3}\left(1+r^{2}+r^{4}\right)$
$=5(1+2+4)=35$