If three geometric means be inserted between 2 and 32, then the third geometric mean will be

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Solution

The correct option is C 16
$2, g_{1}, g_{2}, g_{3},$ 32 where
$a = 2, a r = g_{1}, a r^{2} = g_{2}, a r^{3} = g_{3} a n d a r^{4} = 32$
Now $2 \times r^{4} = 32 \Rightarrow r^{4} = 16 = (2)^{4} = 16 = (2)^{4} \Rightarrow r = 2.$
Then third geometric mean $a r^{2} = 2 \times 2^{3} = 16.$

Aliter : By formula, $G_{3} = 2 . {(\frac{32}{2})}^{3 / 4} = 2.8 = 16.$

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