a, b, c are in G.P. a + b + c = xb, x cannot be
(a) 2
(b) – 2
(c) 3
(d) 4

Hint: Here, use the property of G.P. First represent a, b and c in standard geometric sequence terms by taking common ratio as r. Then compare both sides of equations given in question. By simplifying the equation you will get a quadratic equation, using the concept of real values of r in the quadratic equation as we know r is real. After getting values of x compare the values with the option to get the answer.

Complete step-by-step answer:
 Given terms a, b, c are in G.P. Let r be the common ratio of the G.P.
Then, b = ar and c =\[a{r^2}\]
Now, a + b + c = xb
$\Rightarrow$ a + ar + \[a{r^2}\] = xar
  $\Rightarrow {r^2}$ + (1 − x)r + 1 = 0 This is the quadratic equation with variable r.
But, r is real.
∴ Discriminant > 0
$\Rightarrow {(1 - x)^2} - 4 \times 1 \times 1 = 0$
 $\Rightarrow {x^2}$ − 2x – 3 > 0 $\Rightarrow {x^2}$ − 3x + x – 3 > 0
$\Rightarrow$ x(x − 3) + 1 (x − 3) > 0 $\Rightarrow$ (x − 3)(x + 1) > 0
$\Rightarrow$ x < −1 or x > 3
So, x cannot be 2 and 3.

So, the correct answer is “Option A AND C”.

Note: In these types of questions, use the concept of G.P. Also remember that all terms of G.P. are always real. So if you get any quadratic equation while solving, always consider the equation for real values. Always change the terms of G.P. by assuming common ratio as r if not given, do not for direct calculation without changing as this makes your calculations complicated.