Question

If $a,b,c$ are three distinct real numbers in $G.P.$ and $a+b+c=xb$, then prove that either $x<-1$ or $x>3$.

Answer 1

The given relation implies $a(1+r+r^2)=x\left(ar\right)$
or $r^2+(1-x)r+1=0$, $r$ is real
$\therefore (1-x{)}^2-4\ge 0$ or $x^2-2x-3=+ive$
or $(x+1)(x-3)is+ive$.
$\therefore$ either $x<-1$ or $x>3$.