Question

Find four numbers forming a GP in which the third term is greater than the first term by 9 and the second term is greater then $4^{th}$ by 18.

• A. 2 , -6, 12, -24
• B. 3 , 6, 12 , 24
• C. 3 , -6, 12 , -24
• D. None of these

Answer 1

The correct option is C 3 , -6, 12 , -24

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

$a_1=a,a_2=ar,a_3=a{r}^2,a_4=a{r}^3$

By the given condition,

$a_3=a_1+9$

$\Rightarrow a{r}^2=a+\mathrm{9....}\left(1\right)$

$a_2=a_4+18$

$\Rightarrow ar=a{r}^3+\mathrm{18....}\left(2\right)$

From (1) and (2), we get

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

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

Dividing (4) and (3), we get

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

$\Rightarrow -r=2$

$\Rightarrow r=-2$

Substituting the value of r in (1), we get

$4a=a+9$

$\Rightarrow 3a=9$

$\therefore a=3$

$\Rightarrow a_1=3$

$\Rightarrow a_2=3(-2)=-6$

$\Rightarrow a_3=3(-2{)}^2=12$

$\Rightarrow a_4=3(-2{)}^3=-24$

Thus the first four numbers of the G.P are 3,-6,12 and -24.