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nth Term of HP

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Question

Let $a$ and $b$ be two positive real numbers. Suppose $A_{1}, A_{2}$ are two arithmetic means; $G_{1}, G_{2}$ are two geometric means and $H_{1}, H_{2}$ are two harmonic means between $a$ and $b$ , then-

A

\frac{G_{1} G_{2}}{H_{1} H_{2}} = \frac{A_{1} + A_{2}}{H_{1} + H_{2}}

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B

\frac{G_{1} G_{2}}{H_{1} H_{2}} - \frac{5}{9} = \frac{2}{9} (\frac{a}{b} + \frac{b}{a})

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C

\frac{H_{1} + H_{2}}{A_{1} + A_{2}} = \frac{9 a b}{(2 a + b) (a + 2 b)}

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D

\frac{G_{1} G_{2}}{H_{1} H_{2}} = \frac{H_{1} + H_{2}}{A_{1} + A_{2}}

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Solution

The correct options are
A $\frac{G_{1} G_{2}}{H_{1} H_{2}} = \frac{A_{1} + A_{2}}{H_{1} + H_{2}}$
B $\frac{G_{1} G_{2}}{H_{1} H_{2}} - \frac{5}{9} = \frac{2}{9} (\frac{a}{b} + \frac{b}{a})$
D $\frac{H_{1} + H_{2}}{A_{1} + A_{2}} = \frac{9 a b}{(2 a + b) (a + 2 b)}$
$A_{1} = a + \frac{1}{3} (b - a), A_{2} = b + \frac{2}{3} (b - a)$
$\Rightarrow A_{1} + A_{2} = a + b$
Similarly, $G_{1} = a (\frac{b}{a})^{1 / 3}, G_{2} = a (\frac{b}{a})^{2 / 3}$
$\Rightarrow G_{1} G_{2} = a b$
and
$\frac{1}{H_{1}} = \frac{1}{a} + \frac{1}{3} (\frac{1}{b} - \frac{1}{a})$ ,
$\frac{1}{H_{2}} = \frac{1}{a} + \frac{2}{3} (\frac{1}{b} - \frac{1}{a})$
Now, $\frac{1}{H_{1}} + \frac{1}{H_{2}} = \frac{1}{a} + \frac{1}{b}$
$\Rightarrow \frac{H_{1} + H_{2}}{H_{1} H_{2}} = \frac{a + b}{a b} = \frac{A_{1} + A_{2}}{G_{1} G_{2}}$
$\Rightarrow \frac{G_{1} G_{2}}{H_{1} H_{2}} = \frac{A_{1} + A_{2}}{H_{1} + H_{2}}$
Now, $H_{1} + H_{2} = \frac{3 a b}{a + 2 b} + \frac{3 a b}{2 a + b} = \frac{9 a b (a + b)}{(a + 2 b) (2 a + b)}$
$\Rightarrow \frac{A_{1} + A_{2}}{H_{1} + H_{2}} = \frac{2 (a^{2} + b^{2}) + 5 a b}{9 a b}$
Thus, $\frac{G_{1} G_{2}}{A_{1} A_{2}} - \frac{5}{9} = \frac{2}{9} (\frac{a}{b} + \frac{b}{a})$
Hence, option A, B and C are correct.

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