Question

Find the 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 $4^{th}$ by $18$.

Answer 1

Let $a$ be the first term and $r$ is the common ratio of the G.P.
$a_1=a,a_2=ar,a_3={ar}^2,a_4={ar}^3$
By the given condition $a_3=a_1+9$
$\Rightarrow {ar}^2=a+9...\left(1\right)a_2=a_4+18\Rightarrow ar={ar}^3+18...\left(2\right)$
From (1) and (2) , we obtain $a(r^2-1)=9....\left(3\right)ar(1-r^2)=18....\left(4\right)$
Dividing (4) by (3), we obtain
$\frac{ar(1-r^2)}{a(r^2-1)}=\frac{18}{9}\Rightarrow -r=2\Rightarrow r=-2$
Substituting the value of $r$ in (1), we obtain $4a=a+9\Rightarrow 3a=9\therefore a=3$
Thus, the first four numbers of the G.P. are $3,3(-2),3{(-2)}^2$ i.e.$3,-6,12,$ and $-24$