Question

If the product of three consecutive terms of G.P. is $216$ and the sum of product of pair-wise is $156$, then the numbers will be

• A. $1,3,9$
• B. $2,6,18$
• C. $3,9,27$
• D. $2,4,8$

Answer 1

The correct option is B

$2,6,18$

Explanation of the correct option.

Step $1$: Find the equation for $r$.

Let $\frac{a}{r},a,ar$ be the $3$ consecutive terms of G.P.

Given: Product of three consecutive terms of G.P. is $216$.

$$\begin{gathered} \left(\frac{a}{r}\right)a\left(ar\right)=216 \\ \Rightarrow a^3=216 \\ \Rightarrow a=6 \end{gathered}$$

Given : Sum of the product of pair-wise is $156$.

$$\begin{gathered} \left(\frac{a}{r}\right)a+a\left(ar\right)+ar\left(\frac{a}{r}\right)=156 \\ \Rightarrow a^2\left[\left(\frac{1}{r}\right)+r+1\right]=156 \\ \Rightarrow 36\left(r^2+r+1\right)=156r \\ \Rightarrow r^2+r+1=\left(\frac{156}{36}\right)r \\ \Rightarrow r^2+r+1=\left(\frac{13}{3}\right)r \\ \Rightarrow 3r^2-10r+3=0 \end{gathered}$$

Step $2$: Find the value of $r$.

The quadratic equation is $3r^2-10r+3=0$

$$\begin{gathered} \Rightarrow r=\frac{10\pm \sqrt{100-36}}{6} \\ \Rightarrow r=\frac{10\pm 8}{6} \end{gathered}$$

Thus the value of $r$ is $3$ or $\frac{1}{3}$.

So the numbers are $2,6,18$ or $18,6,2$.

Hence option $\left(B\right)$ is the correct option.