Between 4 and 2916, if (2n+1) G.M.s are inserted, then what is the value of the (n+1)th G.M. ?A) 36​B) 54​C) 108​D) 324

Dear student 

G1, G2, G3----------- Gn+1, ------- G2n, G2n+1, 2916 are in G.P. So Gn+1 will be the middle mean of (2n+1) odd means and so it will be equidistant from 1st and last term.
So, 4, Gn+1, 2916 are also in G.P.

And thus,
Gn+1 = (4.2916)^1/2
= 108 
Hope this hhelps