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Let a, b be...

Question

Let $a, b$ be positive real numbers. If $a, G_{1}, G_{2}, b$ are in geometric progression and $a, H_{1}, H_{2}, b$ are in harmonic progression, show that
$\frac{G_{1} G_{2}}{H_{1} H_{2}} = = \frac{(2 a + b) (a + 2 b)}{9 a b}$

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Solution

Now a, $H_{1}, H_{2}$ ,b are in H.P.
$∴ \frac{1}{a}, \frac{1}{H_{1}}, \frac{1}{H_{2}}, \frac{1}{b}$ are in A.P.
$∴ \frac{1}{b} = \frac{1}{a} + 3 D$ $∴ D = \frac{1}{3} (\frac{1}{b} - \frac{1}{a})$
$∴ \frac{1}{H_{1}} = \frac{1}{a} + D = \frac{1}{a} + \frac{1}{3} (\frac{1}{b} - \frac{1}{a}) = \frac{2 b + a}{3 a b}$
$∴ \frac{1}{H_{2}} = \frac{1}{a} + 2 D = \frac{1}{a} + \frac{2}{3} (\frac{1}{b} - \frac{1}{a}) = \frac{2 a + b}{3 a b}$
$∴ \frac{G_{1} G_{2}}{H_{1} H_{2}} = \frac{a b (2 b + a) (2 a + b)}{9 a^{2} b^{2}} = \frac{(2 b + a) (2 a + b)}{9 a b}$
Note : The value of $H_{2}$ can be obtained from the value of $H_{1}$ if we interchange a and b.

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