Question

If the product of three terms of G.P. is $512$. If $8$ added to the first and $6$ added to the second term, so that number may be in A.P., then the numbers are

• A. $2,4,8$
• B. $4,8,16$
• C. $3,6,12$
• D. None of these

Answer 1

The correct option is B

$4,8,16$

Explanation of the correct option.

Step $1$: Find the value of $a$.

Let $\frac{a}{r},a,ar$ be the terms in GP.

Given : The product of three terms of G.P. is $512$.

Thus, $\left(\frac{a}{r}\right)a\left(ar\right)=512$

$$\begin{gathered} \Rightarrow a^3=512 \\ \Rightarrow a=8 \end{gathered}$$

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

Given : If $8$ added to first and $6$ added to the second term, then numbers are in A.P.

Thus, $\frac{a}{r}+8,a+6,ar$ are in A.P.

So,

$$\begin{gathered} 2\left(a+6\right)=\frac{a}{r}+8+ar \\ \Rightarrow 2\left(8+6\right)=\frac{8}{r}+8+8r \\ \Rightarrow 28=8\left(\frac{1}{r}+1+r\right) \\ \Rightarrow \frac{7}{2}=\frac{1}{r}+1+r \\ \Rightarrow \frac{1}{r}+r-\frac{5}{2}=0 \\ \Rightarrow r^2-\frac{5}{2}r+1=0 \\ \Rightarrow 2r^2-5r+2=0 \\ \Rightarrow 2r^2-4r-r+2=0 \\ \Rightarrow 2r(r-2)-1(r-2)=0 \\ \Rightarrow \left(2r-1\right)\left(r-2\right)=0 \end{gathered}$$

The value of $r$ is $\frac{1}{2}$ or $2$.

Therefore, the terms are $16,8,4$ or $4,8,16$.

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