Question

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 by $4^{th}$ by a18.

Answer 1

Let the four numbers in G.P. be $a,ar,a{r}^2,a{r}^3$

Thus, $a{r}^2=a+9andar=a{r}^3+18$

$\Rightarrow a{r}^2-a=9andar-a{r}^3=18$

Now~ $a{r}^2-a=9\Rightarrow a_r^2-1=9....\left(1\right)$

and~ $ar-a{r}^3=18\Rightarrow ar(1-r^2)\Rightarrow 18$

$\Rightarrow -ar(r^2-1)=18$ ....(2)

Dividing (2) by (1), we have :

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

Now, substituting r in (1), we have :

$a(4-1)=9\Rightarrow a=3$

$ar=3\times (-2)=-6$

$a{r}^2=3\times (-2{)}^2=12$

$a{r}^3=3\times (-2{)}^3=-24$

Thus, required numbers are 3, -6, 12, -24.