Question

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

• A. $4$
• B. $-3$
• C. $-2$
• D. $2$

Answer 1

The correct option is D $2$

$\frac{b}{r},b,br\to G.P.(|r|\ne 1)$

given $a+b+c=xb$

$\Rightarrow \frac{b}{r}+b+br=xb$

$\Rightarrow b=0$ (not possible)

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

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

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

So $x$ can't be ${}^{\prime }{2}^{\prime }$.