Question

Let $a,b,c$ be three distinct real numbers in geometric progression. If $x$ is real and $a+b+c=xb,$ then $x$ can be

• A. $-2$
• B. $3$
• C. $-1$
• D. $4$

Answer 1

Answer: (A), (D)

$4$

Let the common ratio be $r.$
Then $a(1+r+r^2)=xar$
$\Rightarrow 1+r+r^2=xr$
$\Rightarrow r^2+(1-x)r+1=0$
$\Delta \ge 0$
$\Rightarrow (1-x{)}^2-4\ge 0$
$\Rightarrow (1-x+2)(1-x-2)\ge 0$
$\Rightarrow (x-3)(x+1)\ge 0$
$\Rightarrow x\le -1$ or $x\ge 3$

When $x=-1,$
then $r^2+2r+1=0$
$\Rightarrow (r+1{)}^2=0$
$\Rightarrow r=-1$
But then $a=c,$ which is a contradiction.

When $x=3,$
then $r^2-2r+1=0$
$\Rightarrow (r-1{)}^2=0$
$\Rightarrow r=1$
But then $a=b=c,$ which is a contradiction.

Hence, $x<-1$ or $x>3$