Question

If the product of three terms of GP is 512. If 8 added to the first term and 6 added to the second term, so that the numbers may be in AP, then the numbers are?
(A) 2, 4, 8
(B) 4, 8, 16
(C) 3, 6, 12
(D) None of these

Answer 1

Hint: Assume that the three terms of the GP are $a$, $ar$ and $a{{r}^{2}}$ where ‘a’ is the first term and ‘r’ is the common ratio. Take the product of these terms and equate them with 512 to form a relation between a and r. Now, add 8 with $a$ and 6 with $ar$, consider the terms \[\left( a+8 \right)\], $\left( ar+6 \right)$, $a{{r}^{2}}$ in AP and use the formula for the arithmetic mean given as $2\left( ar+6 \right)=\left( a+8 \right)+\left( a{{r}^{2}} \right)$ to form the second relation between a and r. Solve the two equations for the values of a and r, substitute the values in $a$, $ar$ and $a{{r}^{2}}$ to get the numbers.

Complete step-by-step solution:
Here we have been provided with a GP whose product of three terms is 512 and we are adding 8 to the first term and 6 to the second term such that the new numbers form an AP. We are asked to find the three numbers in GP.
Now, three numbers in G.P are assumed as $a$, $ar$ and $a{{r}^{2}}$ where ‘a’ is the first term and ‘r’ is the common ratio. In the first case it is said that the product of these three terms is 512, so we have,
$\begin{align}
  & \Rightarrow a\times ar\times a{{r}^{2}}=512 \\
 & \Rightarrow {{a}^{3}}{{r}^{3}}=512 \\
 & \Rightarrow {{\left( ar \right)}^{3}}=512 \\
\end{align}$
Taking cube root both sides we get,
$\begin{align}
  & \Rightarrow \left( ar \right)=\sqrt[3]{512} \\
 & \Rightarrow \left( ar \right)=8.......\left( i \right) \\
\end{align}$
In the second case 8 is added to the first term and 6 is added to the second term so that we get the new numbers in AP. That means the terms \[\left( a+8 \right)\], $\left( ar+6 \right)$, $a{{r}^{2}}$ are in AP. We know that if three numbers x, y, z are in AP then the arithmetic mean relation is given as 2y = x + z. So, for the new numbers we have,
$\Rightarrow 2\left( ar+6 \right)=\left( a+8 \right)+\left( a{{r}^{2}} \right)$
Substituting the value of r in terms of a from equation (i) we get,
$\begin{align}
  & \Rightarrow 2\left( 8+6 \right)=\left( a+8 \right)+\left( a\times \dfrac{64}{{{a}^{2}}} \right) \\
 & \Rightarrow 28=a+8+\dfrac{64}{a} \\
 & \Rightarrow a-20+\dfrac{64}{a}=0 \\
\end{align}$
Multiplying both the sides with a we get,
$\Rightarrow {{a}^{2}}-20a+64=0$
Using the middle term split method to factorize the above quadratic equation we get,
$\begin{align}
  & \Rightarrow {{a}^{2}}-16a-4a+64=0 \\
 & \Rightarrow \left( a-16 \right)\left( a-4 \right)=0 \\
\end{align}$
Substituting each term equal to 0 one by one we get,
$\Rightarrow \left( a-16 \right)=0$ or $\left( a-4 \right)=0$
$\Rightarrow a=16$ or $a=4$
Case (i): - Considering $a=16$ we have,
$\begin{align}
  & \Rightarrow r=\dfrac{8}{16} \\
 & \Rightarrow r=\dfrac{1}{2} \\
\end{align}$
Therefore, the three numbers in GP will be given as: -
$\Rightarrow a=16$
$\Rightarrow ar=8$
$\Rightarrow a{{r}^{2}}=4$
Case (ii): - Considering $a=4$ we have,
$\begin{align}
  & \Rightarrow r=\dfrac{8}{4} \\
 & \Rightarrow r=2 \\
\end{align}$
Therefore, the three numbers in GP will be given as: -
$\Rightarrow a=4$
$\Rightarrow ar=8$
$\Rightarrow a{{r}^{2}}=16$
From the above two cases we can conclude that the three numbers in G.P are 4, 8, 16. Hence, option (B) is the correct answer.

Note: Note that we have obtained two cases for the values of ‘a’ and ‘r’ because in the first case the G.P is increasing as the value of r is greater than 1 while in the second case the G.P is decreasing as the value of r is less than 1. However, there will be only one set of the three numbers in GP and that can be written in any order.