Question

The sum of first three terms of a $G.P.$ is $\frac{7}{9}$ and their product is $-\frac{8}{27}$ . Find the common ratio of the series

• A. $r=-2/3$ or $-3/2$
• B. $r=-2/3$ or $3/2$
• C. $r=2/3$ or $-3/2$
• D. $r=2/3$ or $3/2$

Answer 1

Answer: (A)

Let first three terms of a GP are a, ar, $a{r}^2.$
Then, $sum=\frac{7}{9}$ (given)
$\Rightarrow$ $a+ar+a{r}^2=\frac{7}{9}$
$\Rightarrow$ $a(1+r+r^2)=\frac{7}{9}$ ..(i)
and Product
$=-\frac{8}{27}$
$\Rightarrow$ $a.ar.ar=-\frac{8}{27}$
$\Rightarrow$ $a^3{r}^3=-\frac{8}{27}={\left(\frac{-2}{3}\right)}^3$
On comparing the powers, we get $ar=\frac{-2}{3}$ ..(ii)
On dividing Eq. (i) by (ii), we get $\frac{1+r+r^2}{r}=\frac{\frac{7}{9}}{\frac{-2}{3}}$
$\Rightarrow$ $\frac{1+r+r^2}{r}=-\frac{7}{9}\times \frac{3}{2}=\frac{-7}{6}$
$\Rightarrow$ $1+r+r^2=\frac{-7}{6}r$
$\Rightarrow$ $r^2+\left(1+\frac{7}{6}\right)r+1=0$
$\Rightarrow$ $r^2+\frac{13}{6}r+1=0$
$\Rightarrow$ $\left(r+\frac{3}{2}\right)\left(r+\frac{2}{3}\right)=0$
$\Rightarrow$ $r=\frac{-3}{2}$ or $\frac{-2}{3}$