Question

$A.M$ of the three numbers which are in $G.P$. is $\frac{14}{3}$. If adding $1$ in the first and second number and subtracting $1$ from the third number, the resulting numbers are in $A.P$. then the sum of the squares of the original three numbers is

• A. $91$
• B. $80$
• C. $84$
• D. $88$

Answer 1

The correct option is C

$84$

The explanation for the correct option:

Step 1: Expressing the given data:

Let $a,ar,a{r}^2$ be the numbers in $G.P$

Given

$AM=\frac{14}{3}$

Step 2: Finding the value of r

$\begin{array}{rcl}\frac{a+ar+a{r}^2}{3}& =& \frac{14}{3}\\ a+ar+a{r}^2& =& 14\\ a(r^2+r+1)& =& 14\dots \left(i\right)\end{array}$

$a+1,ar+1,a{r}^2-1$ are in AP.

So

$\begin{array}{rcl}2(ar+1)& =& a+1+a{r}^2-1\\ 2ar+2& =& a+a{r}^2\\ a{r}^2-2ar+a& =& 2\\ a(r^2-2r+1)& =& 2\dots \left(ii\right)\end{array}$

Divide (i) by (ii)

$\begin{array}{rcl}(r^2+r+1)/(r^2-2r+1)& =& 14/2\\ (r^2+r+1)& =& 7(r^2-2r+1)\\ 6r^2-15r+6& =& 0\\ 2r^2-5r+2& =& 0\\ r& =& 2orr=½\end{array}$

Step 4: Finding the value of $a$

Substitute $r=2$, in (i)

$\begin{array}{rcl}a(2^2+2+1)& =& 14\\ a(4+2+1)& =& 14\\ a\left(7\right)& =& 14\\ a& =& 2\end{array}$

Hence, the numbers are, $2,4,8$.

Step 5: Finding the sum of squares

Sum of square $\begin{array}{rcl}& =& 2^2+4^2+8^2\end{array}$

$\begin{array}{rcl}& =& 4+16+64\\ & =& \mathbf{84}\end{array}$

Hence option (3) is the answer.