If a, b and c be three distinct real numbers

$\frac{\mathrm{b}}{\mathrm{r}}, \mathrm{b}, \mathrm{br} \rightarrow$ G.P. $\quad(|\mathrm{r}| \neq 1)$

given $a+b+c=x b$

$\Rightarrow b / r+b+b r=x b$

$\Rightarrow b=0($ not possible $)$

or $1+\mathrm{r}+\frac{1}{\mathrm{r}}=\mathrm{x} \Rightarrow \mathrm{x}-1=\mathrm{r}+\frac{1}{\mathrm{r}}$

$\Rightarrow x-1>2$ or $x-1<-2$

$\Rightarrow x>3 \quad$ or $x<-1$

So x can't be '2'