Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms. If $a_1 a_5=28$ and $a_2+a_4=29$, then $a_6$ is equal to:

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{a}_1 \cdot \mathrm{a}_5=28 \Rightarrow \mathrm{a} \cdot \mathrm{ar}^4=28 \Rightarrow \mathrm{a}^2 \mathrm{r}^4=28 \quad \ldots \text { (1) } \ & a_2+a_4=29 \Rightarrow a r+a r^3=29 \ & \Rightarrow \operatorname{ar}\left(1+\mathrm{r}^2\right)=29 \ & \Rightarrow \mathrm{a}^2 \mathrm{r}^2\left(1+\mathrm{r}^2\right)^2=(29)^2 \quad \ldots \text { (2) } \end{aligned}$ By Eq. (1) & (2) $\begin{aligned} & \frac{r^2}{\left(1+r^2\right)^2}=\frac{28}{29 \times 29} \ & \Rightarrow \frac{r}{1+\mathrm{r}^2}=\frac{\sqrt{28}}{29} \Rightarrow r=\sqrt{28} \ & \because a^2 r^4=28 \Rightarrow a^2 \times(28)^2=28 \ & \Rightarrow \mathrm{a}=\frac{1}{\sqrt{28}} \ & \therefore \mathrm{a}_6=\mathrm{ar}^5=\frac{1}{\sqrt{28}} \times(28)^2 \sqrt{28}=784 \end{aligned}$ *

Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms. If $a_1 a_5=28$ and $a_2+a_4=29$, then…

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