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Question
If one A.M., A and two geometric means G1 and G2 inserted between any two positive numbers, show that \[\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} = 2A\]
Solution
Let the two positive numbers be a and b .
a, A and b are in A . P .
\[ \therefore 2A = a + b . . . . . . . . (i)\]
\[\text { Also }, a, G_1 , G_2 \text { and b are in G . P } . \]
\[ \therefore r = \left( \frac{b}{a} \right)^\frac{1}{3} \]
\[\text { Also, } G_1 = \text { ar and } G_2 = a r^2 . . . . . . . . . (ii)\]
\[\text { Now, LHS } = \frac{{G_1}^2}{G_2} + \frac{{G_2}^2}{G_1}\]
\[ = \frac{\left( ar \right)^2}{a r^2} + \frac{\left( a r^2 \right)^2}{ar} \left[ \text { Using }(ii) \right]\]
\[ = a + a r^3 \]
\[ = a + a \left( \left( \frac{b}{a} \right)^\frac{1}{3} \right)^3 \]
\[ = a + a\left( \frac{b}{a} \right)\]
\[ = a + b\]
\[ = 2A\]
\[ = \text { RHS } \left[ \text { Using } (i) \right]\]
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