Question

$a,b,c$ are real numbers which are in $G.P.a+b+c=xb$. Which is not the value of $x$

• A. $4$
• B. $-3$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Given,

a,b,c are in G.P

$\therefore a=a$

$b=ar$

$c=a{r}^2$

$a+ar+a{r}^2=x\left(ar\right)$

$\Rightarrow 1+r+r^2=x\gamma$

$\Rightarrow r^2+x\to xr+1=0$

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

$\therefore D>0$

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

$\Rightarrow (1-x-2)(1-x+2)>0$

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

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

$\because x\in [-\infty ,-1)\cup (3,\infty )$

Therefore 2 does not lies betn interval of x

$\text{option}c=2$