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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
$30$
$26$
$35$
$32$
Solution
$a, ar, ar ^{2}, \ldots$
$T _{2}+ T _{6}=\frac{25}{2} \Rightarrow \operatorname{ar}\left(1+ r ^{4}\right)=\frac{25}{2}$
$a^{2} r^{2}\left(1+r^{4}\right)^{2}=\frac{625}{4}$ $….(1)$
$T _{3} \cdot T _{5}=25 \Rightarrow\left( ar ^{2}\right)\left( ar ^{4}\right)=25$
$a^{2} r^{6}=25$ $….(2)$
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) $a ^{2} r ^{6}=25 \Rightarrow\left( ar ^{3}\right)^{2}=25$
$T _{4}+ T _{6}+ T _{8}= ar ^{3}+ ar ^{5}+ ar ^{7}$
$=\operatorname{ar}^{3}\left(1+ r ^{2}+ r ^{4}\right)$
$=5(1+2+4)=35$
