Question

If $a$, $b$ and $c$ be three distinct real numbers in G.P. and $a+b+c=xb$, then $x$ cannot be :

• A. -2
• B. -3
• C. 2
• D. 4

Answer 1

The correct option is C 2

Given :
$a$, $b$ and $c$ are in G.P.
$a+b+c=xb$
Let 'r' be the common ratio of the G.P. Then,
$a+b+c=xb$
$\Rightarrow a+ar+a{r}^2=x\left(ar\right)$
Dividing the equation by $a$,
$\Rightarrow 1+r+r^2=xr$
$\Rightarrow \frac{1}{r}+1+r=x$

We know that $r+\frac{1}{r}\ge 2\text{or}r+\frac{1}{r}\le -2$
$\therefore x\ge 3\text{or}x\le -1$
Hence, $x$ cannot be equal to 2.