If {G_1} and {G_2} are two geometric means an | vedclass

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(C) Let the two numbers be $p$ and $q$.Since ${G_1}$ and ${G_2}$ are two geometric means between $p$ and $q$,the sequence $p, G_1, G_2, q$ is in geome

Vedclass · Accepted Answer

$2A$

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(C) Let the two numbers be $p$ and $q$.Since ${G_1}$ and ${G_2}$ are two geometric means between $p$ and $q$,the sequence $p, G_1, G_2, q$ is in geometric progression.Let the common ratio be $r$. Then $G_1 = pr$,$G_2 = pr^2$,and $q = pr^3$,so $r = (q/p)^{1/3}$.Thus,${G_1} = p(q/p)^{1/3} = p^{2/3}q^{1/3}$ and ${G_2} = p(q/p)^{2/3} = p^{1/3}q^{2/3}$.The arithmetic mean $A$ is given by $A = \frac{p+q}{2}$,so $p+q = 2A$.Now,$\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}} = \frac{(p^{2/3}q^{1/3})^2}{p^{1/3}q^{2/3}} + \frac{(p^{1/3}q^{2/3})^2}{p^{2/3}q^{1/3}} = \frac{p^{4/3}q^{2/3}}{p^{1/3}q^{2/3}} + \frac{p^{2/3}q^{4/3}}{p^{2/3}q^{1/3}} = p + q$.Substituting $p+q = 2A$,we get the value as $2A$.

If ${G_1}$ and ${G_2}$ are two geometric means and $A$ is the arithmetic mean inserted between two numbers,then the value of $\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}}$ is

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