Thank you for reporting, we will resolve it shortly
Q.
If $m$ is the AM of two distinct real numbers $l$ and $n (l, n > 1)$ and $G_1, G_2$ and $G_3$ are three geometric between $l$ and $n$, then $G_1^4 +2G_2^4 +G_3^4$ equals
Given, $m$ is the AM of $l$ and $n$.
$\therefore l + n = 2m ...(i)$
and $G_1 , G_2,G_3$ are geometric means between $l$ and $n$.
$ l, G_1 , G_2,G_3 $ are in GP.
Let $r$ be the common ratio of this GP.
$\therefore G_1 = lr , G_2 = lr^2 , G_3 = lr^3,n-lr^4\, \Rightarrow r =\bigg(\frac{n}{l}\bigg)^{\frac{1}{4}}$
Now,$ G_1^4 + 2G_2^4 + G^4_3 =(lr)^4 + 2(lr^2)^4 +(lr^3)^4$
$ =l^4 \times r^4 (1+ 2r^4 +r^6) =l^4 \times r^4 +(r^4+1)^2$
$ =l^4 \times \frac{n}{l} \bigg(\frac{n+l}{l}\bigg)^2 =ln \times 4m^2 = 4lm^2n$