Question

$a,b,c$ are in $G.P.a+b+c=xb,x$ can not be

• A. $2$
• B. $-2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $2$

if $a,b,c$ are in GP then

$a=a$

$b=ar$

$c=a{r}^2$

Given that $a+b+c=xb$

After substituting the valves we get $\Rightarrow a+ar+a{r}^2=x\left(ar\right)$

$\Rightarrow a(1+r+r^2)=a\left(xr\right)$

$\Rightarrow 1+r+r^2=xr$

$\Rightarrow x=\frac{r^2+r+1}{r}=r+1+\frac{1}{r}$

we know that, $r+\frac{1}{r}⩾2$ if $r>0$

$r+\frac{1}{r}⩽-2\text{if}r<0$

Therefore

$2⩽r+\frac{1}{r}<\infty$

$3⩽r+1+\frac{1}{r}<\infty$

$3⩽x<\infty$ for $r>0$

for $r<0$

$-\infty <r+\frac{1}{r}⩽-2$

$-\infty <r+\frac{1}{r}+1⩽-1$

$-\infty <x⩽-1$

$\therefore 2$ is not the possible value of $x$ Hence $\left(A\right)$ is the correct option.