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} $