Question

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

• A. $r=-\frac{2}{3}$or $-\frac{3}{2}$
• B. $r=-\frac{2}{3}$or $\frac{3}{2}$
• C. $r=\frac{2}{3}$or $-\frac{3}{2}$
• D. $r=\frac{2}{3}$or $\frac{3}{2}$

Answer 1

The correct option is A

$r=-\frac{2}{3}$or $-\frac{3}{2}$

Explanation for correct option:

Let the first three terms be $\frac{a}{r},a,ar$

It is given that the product of the first three terms is $-\frac{8}{27}$

$$\begin{gathered} \Rightarrow \frac{a}{r}·a·ar=-\frac{8}{27} \\ \Rightarrow a^3=-\frac{8}{27} \\ \Rightarrow a=-\frac{2}{3} \end{gathered}$$

It is given that the sum of the first three terms is $\frac{7}{9}$

$$\begin{gathered} \Rightarrow \frac{a}{r}+a+ar=\frac{7}{9} \\ \Rightarrow a\left(\frac{1}{r}+1+r\right)=\frac{7}{9} \\ \Rightarrow a\left(\frac{1+r+r^2}{r}\right)=\frac{7}{9} \\ \Rightarrow \frac{1+r+r^2}{r}=-\frac{21}{18}\left[\because a=-\frac{2}{3}\right] \\ \Rightarrow 18+18r+18r^2=-21r \\ \Rightarrow 6r^2+13r+6=0 \end{gathered}$$

Solving the quadratic equation we get

$$\begin{gathered} \Rightarrow r=\frac{-13\pm \sqrt{{13}^2-4·6·6}}{2·6} \\ \Rightarrow r=\frac{-13\pm 5}{12} \\ \Rightarrow r=-\frac{3}{2}\mathrm{or}-\frac{2}{3} \end{gathered}$$

Hence, option A is correct.