Question

In a G.P., if third term is greater than the first term by 9 , and the second term is greater than the fourth term by 18 , then the first four terms of this G.P. are

• A. $3,6,12,24$
• B. $-3,6,-12,24$
• C. $3,-6,12,-24$
• D. $-3,-6,-12,-24$

Answer 1

Answer: (C)

Let the G.P. is $a,ar,a{r}^2,a{r}^3,\dots$
Given, third term $=$ First term $+9$
i.e., $T_3=a+9$
$\Rightarrow a{r}^2=a+9$
$\Rightarrow a{r}^2-a=\mathrm{9....}$(i)
Again, second term $=$ Fourth term $+18$
i.e.,$T_2=T_4+18$
$\Rightarrow ar=a{r}^3+18$
$\Rightarrow ar-a{r}^3=\mathrm{18....}$(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{a{r}^2-a}{ar-a{r}^3}=\frac{9}{18}$
$\Rightarrow \frac{a\left(r^2-1\right)}{ar\left(1-r^2\right)}=\frac{1}{2}$
$\Rightarrow \frac{-1}{r}\frac{\left(1-r^2\right)}{\left(1-r^2\right)}=\frac{1}{2}$
$\Rightarrow -\frac{1}{r}=\frac{1}{2}\Rightarrow r=-2$
Putting $r=-2$ in Eq. (ii), we get
$a(-2)-a(-2{)}^3=18$
$\Rightarrow -2a+8a=18$
$\Rightarrow 6a=18$
$\Rightarrow a=3$
G.P. is $3,3(-2),3(-2{)}^2,3(-2{)}^3,\dots \dots$
i.e., $3,-6,12,-24,\dots \dots$