Mathematics - Sequence & Series Question with Solution | TestHub

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