Let $a, b$and $c$ be distinct real numbers. If $a, b, c$ are in geometric progression and $a + b + c = xb,$ then $x$ lies in the set

$a+b+c=xb$

$\Rightarrow \frac{b}{r}+b+br=xb$, where $r$ is common

ratio of the G.P and $r \ne 1$

$\Rightarrow r+\frac{1}{r}=(x-1)$

But $r+\frac{1}{r}<-2$ or $r+\frac{1}{r}>2$

$\begin{array}{l}\therefore x-1<-2\text{ or }x-1>2\\ \Rightarrow x<-1\text{ or }x>3\\ \Rightarrow x\in (-\mathrm{\infty },-1)\cup (3,\mathrm{\infty })\end{array}$