JEE Main 2024 — 8 April, Shift 2 — Question 4

$\quad T_{2}+T_{6}=\frac{70}{3}$

$\operatorname{ar}+\mathrm{ar}^{5}=\frac{70}{3}$

$\mathrm{T}_{3} \cdot \mathrm{~T}_{5}=49$

$\mathrm{ar}^{2} \cdot \mathrm{ar}^{4}=49$

$a^{2} r^{6}=49$

$\mathrm{ar}^{3}=+7, \mathrm{a}=\frac{7}{\mathrm{r}^{3}}$

$\operatorname{ar}\left(1+\mathrm{r}^{4}\right)=\frac{70}{3}$

$\frac{7}{\mathrm{r}^{2}}\left(1+\mathrm{r}^{4}\right)=\frac{70}{3}, \mathrm{r}^{2}=\mathrm{t}$

$\frac{1}{\mathrm{t}}\left(1+\mathrm{t}^{2}\right)=\frac{10}{3}$

$3 t^{2}-10 t+3=0$

$\mathrm{t}=3, \frac{1}{3}$

Increasing G.P. $r^{2}=3, r=\sqrt{3}$

$\mathrm{T}_{4}+\mathrm{T}_{6}+\mathrm{T}_{8}$

$=\mathrm{ar}^{3}+\mathrm{ar}^{5}+\mathrm{ar}^{7}$

$=\operatorname{ar}^{3}\left(1+\mathrm{r}^{2}+\mathrm{r}^{4}\right)$ $=7(1+3+9)=91$.