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

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(C) Given $m = \frac{l+n}{2}$,so $2m = l+n$.Since $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,the sequence $l, G_1, G_2, G_3, n$ is

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$4lm^2n$

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(C) Given $m = \frac{l+n}{2}$,so $2m = l+n$.Since $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,the sequence $l, G_1, G_2, G_3, n$ is in $G.P.$Let the common ratio be $r$. Then $n = l r^4$,so $r^4 = \frac{n}{l}$.The terms are $G_1 = lr, G_2 = lr^2, G_3 = lr^3$.We need to evaluate $G_1^4 + 2G_2^4 + G_3^4 = (lr)^4 + 2(lr^2)^4 + (lr^3)^4$.$= l^4r^4 + 2l^4r^8 + l^4r^{12} = l^4r^4(1 + 2r^4 + r^8) = l^4r^4(1 + r^4)^2$.Substitute $r^4 = \frac{n}{l}$:$= l^4 \left(\frac{n}{l}\right) \left(1 + \frac{n}{l}\right)^2 = l^3n \left(\frac{l+n}{l}\right)^2 = l^3n \frac{(l+n)^2}{l^2} = \ln(l+n)^2$.Since $l+n = 2m$,we have $\ln(2m)^2 = 4lm^2n$.

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

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