Let $a_1, a_2, a_3, \ldots$ be a G. P. of increasing positive numbers. If $\mathrm{a}_3 \mathrm{a}_5=729$ and $\mathrm{a}_2+\mathrm{a}_4=\frac{111}{4}$, then $24\left(a_1+a_2+a_3\right)$ is equal to

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Aicharya · Accepted Answer

Let the $\mathrm{I}^{	ext {st }}$ term of G.P. be a & common ratio be r $\begin{aligned} \mathrm{a}_3 \mathrm{a}_5 & =\operatorname{ar}^2 \cdot \operatorname{ar}^4=729 \ & =\mathrm{a}^2 \mathrm{r}^6=729 \ & =\mathrm{ar}^3=27 \quad ....(i) \end{aligned}$ $\begin{aligned} \mathrm{a}_2+\mathrm{a}_4 & =\mathrm{ar}+\mathrm{ar}^3=\frac{111}{4} \ & =\mathrm{ar}=\frac{3}{4} \quad ....(ii)\end{aligned}$ $	ext { (i) } \div 	ext { (ii) }$ $\begin{aligned} & \frac{\mathrm{ar}^3}{\mathrm{ar}}=\frac{27}{3 / 4} \ & \mathrm{r}^2=36 \ & \mathrm{r}=6 \end{aligned}$ from (ii) $a(6)=\frac{3}{4} \Rightarrow a=\frac{1}{8}$ Now, $24\left(a_1+a_2+a_3ight)$ $\begin{aligned} & =24\left(a+a r+a^2ight) \ & =24 a\left(1+r+r^2ight) \end{aligned}$ $\begin{aligned} & =24 	imes \frac{1}{8}(1+6+36) \ & =3(43) \ & =129\end{aligned}$ ~

Let $a_1, a_2, a_3, \ldots$ be a G. P. of increasing positive numbers. If $\mathrm{a}_3 \mathrm{a}_5=729$…

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