Between two unequal numbers, if $a_{1}$ , $a_{2}$ are two AMs; $g_{1}$ , $g_{2}$ are two GMs and $h_{1}$ , $h_{2}$ are two HMs then $g_{1}$ , $g_{2}$ equal to

a_{1} h_{1}

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a_{1} h_{2}

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a_{2} h_{2}

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a_{2} h_{1}

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Solution

The correct options are
A $a_{2} h_{1}$
B $a_{1} h_{2}$
Given that, $a, a_{1}, a_{2}, b$ are in A.P
Therefore, $\frac{b - a}{2 + 2 - 1} = a_{1} - a = b - a_{2}$
$\frac{b - a}{2 + 2 - 1} = a_{1} - a = b - a_{2}$ and $a_{2} = \frac{2 b + a}{3}$ .....(1)
and $a, g_{1}, g_{2}, b$ are in G.P
$g_{1} g_{2} = a b$ .....(2)
and $g_{1} g_{2} = a b$ are in H.P
Therefore, $\frac{1}{a}, \frac{1}{h_{1}}, \frac{1}{h_{2}}, \frac{1}{b}$ are in A.P

Therefore, $\frac{1}{2 + 2 - 1} (\frac{1}{b} - \frac{1}{a}) = \frac{1}{h_{1}} - \frac{1}{a} = \frac{1}{b} - \frac{1}{h_{2}}$
$\Rightarrow h_{1} = \frac{3 a b}{a + 2 b}$ and $h_{2} = \frac{3 a b}{2 a + b}$ .........(3)
from (1) & (3), we get

$a_{1} h_{2} = a_{2} h_{1} = a b$
Therefore, from (2), we get
$g_{1} g_{2} = a_{1} h_{2} = a_{2} h_{1}$

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