In a increasing geometric series, the sum of | vedclass

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Question

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

Vedclass · Accepted Answer

$35$

Vedclass · Answer

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

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

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