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If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G₁, G₂ and G₃ are three geometric means between l and n, then $G_{1}^{4}+2G_{2}^{4}+G_{3}^{4} \; equals.$
Option: 1
$4l^2mn$

Option: 2
$4lm^2n$

Option: 3
$4lmn^2$

Option: 4
$4l^2m^2n^2$

Answers (1)

best_answer

As we learnt in

Arithmetic mean of two numbers (AM)

$A=\frac{a+b}{2}$

and

Geometric mean of two numbers (GM) -

$GM= \sqrt{ab}$

Now,

Given l + n = 2m .........(i)

G₁, G₂ and G₃ are three geometric means between l and n

So, n is 5th term of GP: l, G₁, G₂, G₃, n

$\therefore n= l.r^{4}, r=\left ( \frac{n}{l} \right )^\frac{1}{4}$

$\therefore G_{1}^{4}+2\left ( G_{2} \right )^{4}+\left ( G_{3} \right )^{4}= \left ( lr \right )^{4}+2\left ( lr^{2} \right )^{4}+\left ( lr^{3} \right )^{4}$

$= l^{4}r^{4}+2l^{4}r^{8}+l^{4}r^{12}$

$= l^{4}r^{4}\left [ 1+2r^{4}+r^{8} \right ]$

$= l^{4}\cdot \frac{n}{l}\left [ 1+\frac{2n}{l}+\frac{n^{2}}{l^{2}} \right ]$

$=\frac{nl^{3}}{l^{2}}\left [l^{2}+2nl+n^{2} \right ]$

$=nl\left ( l+n \right )^{2}$

$= nl\left ( 2m \right )^{2}$

$= 4m^{2}nl\:\:\: (from \:\text{(i)} )$

Posted by

Divya Prakash Singh

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