Question

$a,b,c$ are three distinct real numbers, which are in G.P. and $a+b+c=xb$, then

• A. $x<-1$
• B. $-1<x<2$
• C. $2<x<3$
• D. $x>3$

Answer 1

Answer: (A), (D)

The correct options are
A $x>3$
D $x<-1$

Given $a,b,c$ are in GP.

Let r be the common ratio.

Then $b=ar,c=a{r}^2$

Also, given $a+b+c=xb$

$\Rightarrow a+ar+a{r}^2=arx$

$\Rightarrow r^2+r(1-x)+1=0$

Since r is real & distinct $\Rightarrow$ D > 0

$\therefore (1-x{)}^2-4>0$

$\Rightarrow x^2-2x-3>0$

$\Rightarrow (x+1)(x-3)>0$

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