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Algebra of Limits

If a be the...

Question

If $a$ be the A.M of two numbers $b$ and $c$ and $G_{1}, G_{2}$ are the two G.Ms between them then prove that $G_{1}^{3} + G_{2}^{3} = 2 a b c$ .

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Solution

We have $G_{1}$ and $G_{2}$ as two geometric means between b and c i.e. $b, G_{1}, G_{2}, c$ are in G.P
$c = b (r)^{4 - 1}$
$c = b r^{3}$
$r = {(\frac{c}{b})}^{\frac{1}{2}}$
$G_{1} = b r = b {(\frac{c}{b})}^{\frac{1}{3}}$
$G_{1}^{3} = b^{3} (\frac{c}{b}) = b^{2} c$
$G_{2} = b r^{2} = b {(\frac{c}{b})}^{\frac{2}{3}}$
$G_{2}^{3} = b r^{2} = b^{3} (\frac{c^{2}}{b^{2}}) = b c^{2}$
As a is AM of b and c, hence
$2 a = b + c$
$G_{1}^{3} + G_{2}^{3} = b^{2} c + b c^{2}$
$= b c (b + c)$
$= 2 b c a$ [Using (i)]
$G_{1}^{3} + G_{2}^{3} = 2 a b c$

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