Question

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

• A. 91
• B. 364
• C. 84
• D. 273

Answer 1

The correct option is C

84

Let the three numbers be $a,ar,a{r}^2$

$\frac{a+ar+a{r}^2}{3}=\frac{14}{3}\Rightarrow a(1+r+r^2)=\mathrm{14...}\left(I\right)$

Also, $a+1,ar+1,a{r}^2-1$ are in A.P.

$\Rightarrow a+1+a{r}^2-1=2(ar+1)$

$\Rightarrow a+a{r}^2=2ar+2$

$\Rightarrow a(r^2-2r+1)=\mathrm{2....}\left(II\right)$

$\left(I\right)÷\left(II\right)\Rightarrow \frac{1+r+r^2}{r^2-2r+1}=\frac{14}{2}=7$

$\Rightarrow 2r^2-5r+2=0\Rightarrow r=2$ or $r=\frac{1}{2}$

$\Rightarrow$ a = 2 or a = 8

Hence required nos. are 2, 4, 8 or 8, 4, 2

Sum of squares = 4 +16 + 64 = 84