JEE Main 2025 — 22 January, Morning Shift — Question 7

$\mathrm{a}_{1} \cdot \mathrm{a}_{5}=28 \Rightarrow \mathrm{a} \cdot \mathrm{ar}^{4}=28 \Rightarrow \mathrm{a}^{2} \mathrm{r}^{4}=28…(1)$

$a_{2}+a_{4}=29 \Rightarrow a r+a r^{3}=29$

$\Rightarrow \operatorname{ar}\left(1+\mathrm{r}^{2}\right)=29$

$\Rightarrow \mathrm{a}^{2} \mathrm{r}^{2}\left(1+\mathrm{r}^{2}\right)^{2}=(29)^{2}…(2)$

By Eq. (1) & (2)

$\frac{r^{2}}{\left(1+r^{2}\right)^{2}}=\frac{28}{29 \times 29}$

$\Rightarrow \frac{r}{1+\mathrm{r}^{2}}=\frac{\sqrt{28}}{29} \Rightarrow r=\sqrt{28}$

$\because a^{2} r^{4}=28 \Rightarrow a^{2} \times(28)^{2}=28$

$\Rightarrow \mathrm{a}=\frac{1}{\sqrt{28}}$

$\therefore \mathrm{a}_{6}=\mathrm{ar}^{5}=\frac{1}{\sqrt{28}} \times(28)^{2} \sqrt{28}=784$