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 $4^{\text {th }}$ by $18 .$

Let a be the first term and r be the common ratio of the $G.P.$

$a_{1}=a, a_{2}=a r, a_{3}=a r^{2}, a_{4}=a r^{3}$

By the given condition,

$a_{3}=a_{1}+9 \Rightarrow a r^{2}=a+9$          ..........$(1)$

$a_{4}=a_{4}+18 \Rightarrow a r=a r^{3}+18$         ..........$(2)$

From $(1)$ and $(2),$ we obtain

$a\left(r^{2}-1\right)=9 $        ..........$(3)$

$a r\left(1-r^{2}\right)=18$          ...........$(4)$

Dividing $(4)$ by $(3),$ we obtain

$\frac{\operatorname{ar}\left(1-r^{2}\right)}{a\left(r^{2}-1\right)}=\frac{18}{9}$

$\Rightarrow-r=2$

$\Rightarrow r=-2$

Substituting the value of $r$ in $(1),$ we obtain

$4 a=a+9$

$\Rightarrow 3 a=9$

$\therefore a=3$

Thus, the first four numbers of the $G.P.$ are $3,3(-2), 3(-2)^{2},$ and $3(-2)^{3}$

i.e., $3,-6,12$ and $-24$