JEE Mains 2019 #1

Let the common ratio of the geometric progression be $r$ . Then $a= \dfrac br$ , $c=br$ , and

$\begin{aligned} a + b + c & = \frac br + b + br = \left(\frac 1r + 1 + r\right)b \\ \implies x & = \frac 1r + 1 + r \end{aligned}$

Multiplying both sides by $r$ and rearranging, $r^2 + (1-x)r + 1 = 0$ $\implies r = \dfrac {x-1 \pm \sqrt{(1-x)^2 - 4}}2$ . For $r$ to be real $(1-x)^2 - 4 \ge 0$ $\implies (1-x)^2 \ge 4$ $\implies x \le -1 \cup x \ge 3$ , and $\boxed 2$ is not in the range.