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  4. (2n + 1) G.M.'s are inserted between 4 and 2916. Then the (n + 1)th G.M. is equal to

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Q. $ (2n + 1) G.M.'s$ are inserted between $4$ and $2916$. Then the $(n + 1)^{th}\, G.M$. is equal to

Sequences and Series

Solution:

The $\left(n+1\right)^{th} G.M$. = middle $G.M. = G.M$. of $4$ and $2916$
$ = \sqrt{4\times2916} = 108$