If m is the A.M of two distinct real numbers | vedclass

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Question

$m=\frac{l+n}{2}$

$\Rightarrow 2 m=l+n$

$G_{1}, G_{2}, G_{3}$

$l, G_{1}, G_{2}, G_{3},n$ are in $GP$

let $d$ be the common ration

$G_{1

Vedclass · Accepted Answer

$4l{m^2}n$

Vedclass · Answer

$m=\frac{l+n}{2}$

$\Rightarrow 2 m=l+n$

$G_{1}, G_{2}, G_{3}$

$l, G_{1}, G_{2}, G_{3},n$ are in $GP$

let $d$ be the common ration

$G_{1}=l d$

$G_{2}=l d^{2}$

$G_{3}=l d^{3}$

$n=l d^{4}$

$G_{1}^{4}+2 G_{1}^{4}+G_{3}^{4}=(l d)^{4}+2\left(l d^{2}\right)^{4}+\left(l d^{3}\right)^{4}$

$=l^{4} d^{4}+2 l^{4} d^{8}+l^{4} d^{12}$

$=l^{4} d^{4}\left[1+2 d^{4}+d^{8}\right]$

$=l^{4} \frac{n}{l}\left[1+2\left(\frac{n}{l}\right)^{4}+\left(\frac{n}{l}\right)^{2}\right]$

$=n l^{3}\left(1+\frac{n}{l}\right)^{2}$

$=n l^{3} \frac{(l+n)^{2}}{l^{2}}$

$=n l(l+n)^{2}$

$=n l(2 m)^{2} a s(2 m=l+n)$

$=4 m^{2} n l$

If $m$ is the $A.M$ of two distinct real numbers $ l$ and $n (l,n>1) $ and $G_1, G_2$ and $G_3$ are three geometric means between $l$ and $n$ then $G_1^4 + 2G_2^4 + G_3^4$ equals :

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