In an increasing geometric progression ol pos | vedclass

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Question

$ \mathrm{T}_2+\mathrm{T}_6=\frac{70}{3} $

$ \mathrm{ar}+\mathrm{ar}^5=\frac{70}{3} $

$ \mathrm{~T}_3 \cdot \mathrm{T}_5=49 $

$ \mathrm{ar}^2

Vedclass · Accepted Answer

$78$

Vedclass · Answer

$ \mathrm{T}_2+\mathrm{T}_6=\frac{70}{3} $

$ \mathrm{ar}+\mathrm{ar}^5=\frac{70}{3} $

$ \mathrm{~T}_3 \cdot \mathrm{T}_5=49 $

$ \mathrm{ar}^2 \cdot \mathrm{ar}^4=49 $

$ \mathrm{a}^2 \mathrm{r}^6=49 $

$ \mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3} $

$ \mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3} $

$ \frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t} $

$ \frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3} $

$ 3 \mathrm{t}^2-10 \mathrm{t}+3=0 $

$ \mathrm{t}=3, \frac{1}{3}$

Increasing $G.P$. $\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}$

$ \mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8 $

$ =\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7 $

$ =\mathrm{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right) $

$ =7(1+3+9)=91$

In an increasing geometric progression ol positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is :-

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