If m is the AM of two distinct real numbers 1 and n1- n-1 and G1- G2 and G3 are three geometric means between 1 and n- then G14-2 G24-G34 equals

The correct option is A 4 lm^2 nGiven, m is the AM of l and n ∴ 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, n ar ...

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Standard XI

Mathematics

Insertion of GP's between 2 Numbers

If m is the A...

Question

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 means between l and n, then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals

4 l^{2} m n

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4 l m^{2} n

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l m n^{2}

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l^{2} m^{2} n^{2}

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Solution

The correct option is B $4 l m^{2} n$
Given, m is the AM of l and n
$∴ l + n = 2 m \dots \dots (i)$
and $G_{1}, G_{2}, G_{3}$ are geometric means between l and n
$l, G_{1}, G_{2}, G_{3}, n$ are in GP
Let r be the common ratio of this GP
$∴ G_{1} = l r, G_{2} = l r^{2}, G_{3} = l r^{3}, n = l r^{4}$
$\Rightarrow r = {(\frac{n}{l})}^{\frac{1}{4}}$
Now, $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4} = (l r)^{4} + 2 (l r^{2})^{4} + (l r^{3})^{4}$
$= l^{4} \times r^{4} (1 + 2 r^{4} + r^{8}) = l^{4} \times r^{4} (r^{4} + 1)^{2}$
$= l^{4} \times \frac{n}{l} {(\frac{n + l}{l})}^{2} = l n \times 4 m^{2} = 4 l m^{2} n$

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If m is the AM 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 G41+2G42+G43 equals

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 means between l and n, then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals