If $m$ is the $A . M$ . of two distinct real numbers $I$ and $n$ $(l, n > 1)$ and $G_{1}, G_{2} and G_{3}$ are three geometric means between $I$ and $n$ , then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals :

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$4 {lmn}^{2}$

$4 l^{2} m^{2} n^{2}$

$4 {lm}^{2} n$

$4 l^{2} mn$

answer is D.

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Detailed Solution

We have, $m = \frac{l + n}{2}$

Let $r$ be the common ratio of the $G.P. l, G_{1}, G_{2}, G_{3}, n$ Then,

$\begin{array}{l} n = l r^{4} \Rightarrow r = {(\frac{n}{l})}^{1 / 4} \\ ∴ G_{1} = l r = l^{3 / 4} n^{1 / 4}, G_{2} = l r^{2} = l^{2 / 4} n^{2 / 4}, G_{3} = l r^{3} = l^{1 / 4} n^{3 / 4} \\ \Rightarrow G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4} = l^{3} n + 2 l^{2} n^{2} + \ln^{3} \end{array}$ $= \ln (l + n)^{2} = \ln (2 m)^{2} = 4 {lm}^{2} n$ .