Question

Between $4$ and $2916,(2n+1)$ G.M's are inserted then $(n+1{)}^{th}$ G.M is

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

Answer 1

The correct option is C $108$

As

$G_1,G_2,G_3,,G_{n+1},....,2916$are in GP

So $G_{n+1}$ is the middle term of (2n+1) odd means.

So It will be equidistant from first and last term.

$\therefore G_{n+1}=(4\times 2916{)}^{\frac{1}{2}}$

$G_{n+1}=108$