Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ be in a geometric progression. $2,7,9,5$ are subtracted respectively from $x_1, x_2, x_3$ $x_4$ then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24}\left(x_1 x_2 x_3 x_4\right)$ is :

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Aicharya · Accepted Answer

$\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4 \rightarrow \text { G.P. }$
Let a, ar, $\mathrm{ar}^2, \mathrm{ar}^3 \rightarrow$ G.P.
Now $a-2, a r-7, a^2-9, \mathrm{ar}^3-5 \rightarrow$ A.P.
$2(\mathrm{ar}-7)=\mathrm{a}-2+\mathrm{ar}^2-9$....(i)
$2\left(\operatorname{ar}^2-9\right)=\mathrm{ar}-7+\mathrm{ar}^3-5$....(ii)
$\begin{aligned} & \text { Solving } \mathrm{r}=2, \mathrm{a}=-3 \\ & \therefore \text { Product }=\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4=\mathrm{a}^4 \mathrm{r}^6=81 \times 64\end{aligned}$
$\begin{aligned} & \text { Solving } \mathrm{r}=2, \mathrm{a}=-3 \\ & \therefore \text { Product }=\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4=\mathrm{a}^4 \mathrm{r}^6=81 \times 64\end{aligned}$ ~

Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ be in a geometric progression. $2,7,9,5$ are…

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