Find four numbers forming a geometric progression in which third term is greater than the first term by 9,

Solution:

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,

$a_{3}=a_{1}+9$

$\Rightarrow a r^{2}=a+9 \ldots(1)$

$a_{2}=a_{4}+18$

$\Rightarrow a r=a r^{3}+18 \ldots(2)$

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

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

$\operatorname{ar}\left(1-r^{2}\right)=18 \ldots(4)$

Dividing (4) by (3), we obtain

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

$\Rightarrow-r=2$

$\Rightarrow r=-2$

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 $-24$.