Question

If the sum of three terms of G.P is $19$ and product is $216$, then the common ratio of the series is

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

Answer 1

The correct option is B

$\frac{3}{2}$

Explanation for the correct option.

Step 1: Form the first equation using the product of the terms.

Let $\frac{a}{r},a,ar$, be the three terms of GP. Now as the product of the terms is $216$, so

$$\begin{gathered} \frac{a}{r}·a·ar=216 \\ \Rightarrow a^3=216 \\ \Rightarrow a=6 \end{gathered}$$

Step 2: Form the second equation using the sum of the terms.

It is given that the sum of the three terms is $19$. So,

$$\begin{gathered} \frac{a}{r}+a+ar=19 \\ \Rightarrow \frac{6+6r+6r^2}{r}=19\left[a=6\right] \\ \Rightarrow 6+6r+6r^2=19r \\ \Rightarrow 6r^2-13r+6=0 \end{gathered}$$

Step 3. Find the common ratio.

Solve the quadratic equation $6r^2-13r+6=0$ using the quadratic formula as:

$\begin{array}{rcl}r& =& \frac{-(-13)\pm \sqrt{{(-13)}^2-4\times 6\times 6}}{2\times 6}\\ & =& \frac{13\pm \sqrt{169-144}}{12}\\ & =& \frac{13\pm \sqrt{25}}{12}\\ & =& \frac{13\pm 5}{12}\end{array}$

So either

$\begin{array}{rcl}r& =& \frac{13+5}{12}\\ & =& \frac{18}{12}\\ & =& \frac{3}{2}\end{array}$

or

$\begin{array}{rcl}r& =& \frac{13-5}{12}\\ & =& \frac{8}{12}\\ & =& \frac{2}{3}\end{array}$

So the value of the common ratio is either $\frac{3}{2}$ or $\frac{2}{3}$.

Hence, the correct option is B.