The sum of two geometric mean's inserted between $3$ and $81$ is

24

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18

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36

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30

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Solution

The correct option is C $36$
Let the required numbers be $G_{1}$ and $G_{2}$ , then
$3, G_{1}, G_{2}, 81$ are in G.P.
Let the common ratio of the G.P be $r$ , we get
$81 = 3 r^{3}$
$\Rightarrow r^{3} = 27 \Rightarrow r = 3$

$∴ G_{1} = 3 r = 9 G_{2} = 3 r^{2} = 27$

Hence, the required sum is $36$ .

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