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 $18 .$
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 $-24$
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