If n geometric means between a and b be $G_{1}, G_{2}, . . . . G_{n}$ and a geometric mean of a and b be G, then the true relation is

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Solution

The correct option is C

Here $G = (a b)^{\frac{1}{2}}$ and $G_{1} = a r^{1}, G_{2} = a r^{2}, . . . . . G_{n} = a r^{n}$
Therefore $G_{1}, G_{2}, G_{3}, . . . . G_{n} = a^{n} r^{1 + 2 + . . + n} = a^{n} r^{\frac{n (n + 1)}{2}}$
But $a r^{n + 1} = b \Rightarrow r = {(\frac{b}{a})}^{\frac{1}{(n + 1)}}$
Therefore, the required product is
$a^{n} {(\frac{b}{a})}^{\frac{1}{(n + 1)} . \frac{n (n + 1)}{2}}$
Note : It is a well known fact.

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