Let $G_1, G_2$ be two numbers between 3 and 81 such that 3, $G_1, G_2$ , 81 is a G.P. Since $a_4$ = 81 ⇒ $ar^3$ = 81 ⇒ $3r^3$ = 81 (Since a = 3) ⇒ $r^3$ = 27 ⇒ r = 3 Since $a^2$ = $G_1$ ⇒ $G_1$ = ar = 3 × 3 = 9 Since $a_3$ = $G_2$ ⇒ $G_2$ = $ar^2$ = 3 × $(3)^2$ = 27 Hence, we can insert 9, 27 between 3 and 81 so that the resulting sequence is in G.P.