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

Answer: (D)

$\frac{b}{r},b,br\to G.P.\left(\left|r\right|\ne 1\right)$
given $a+b+c=xb$
$\Rightarrow$ $b/r+b+br=xb$
$\Rightarrow$ $b=0$ (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 ${}^{\prime }{2}^{\prime }$