Let the four numbers forming a G.P. be a, ar, $ar^2, ar^3$ According to question, $ar^2$ = a + 9 ⇒ $a(r^2 - 1)$ = 9 ....(i) and ar = $ar^3$ + 18 ⇒ ar (1 – $r^2$) = 18 ....(ii) Dividing (i) by (ii), we get $\frac{a(r^2-1)}{ar(1-r^2)}$ = $\frac{9}{18}$ ⇒ $\frac{-1}{r}$ = $\frac{1}{2}$ ⇒ r = - 2 Put r = – 2 in (i), we get a (4 – 1) = 9 ⇒ a = $\frac{9}{3}$ = 3 Hence the four numbers forming a G.P. are 3, – 6, 12, – 24.