Find four numbers forming a geometric progression in which the third term is greater than the first term by $9,$ and the second term is greater than the $4^{\text{th}}$ term by $18.$

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The terms are $a_{1}=a, a_{2}=a r, a_{3}=a r^{2}, a_{4}=a r^{3}.$
By the given conditions:
$a_{3} = a_{1} + 9 \Rightarrow a r^{2} = a + 9$ ..........$(1)$
$a_{2} = a_{4} + 18 \Rightarrow a r = a r^{3} + 18$ ..........$(2)$
From $(1),$ we have $a(r^{2} - 1) = 9.$ ..........$(3)$
From $(2),$ we have $a r(1 - r^{2}) = 18 \Rightarrow -a r(r^{2} - 1) = 18.$ ..........$(4)$
Dividing $(4)$ by $(3):$
$\frac{-a r(r^{2} - 1)}{a(r^{2} - 1)} = \frac{18}{9}$
$-r = 2 \Rightarrow r = -2.$
Substituting $r = -2$ into $(1):$
$a((-2)^{2} - 1) = 9$
$a(4 - 1) = 9$
$3a = 9 \Rightarrow a = 3.$
The four numbers are $a, ar, ar^{2}, ar^{3}.$
Substituting $a = 3$ and $r = -2:$
$3, 3(-2), 3(-2)^{2}, 3(-2)^{3}$
$3, -6, 12, -24.$