Question

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

• A. $36$
• B. $54$
• C. $108$
• D. $324$

Answer 1

Answer: (C)

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