If a, b, c are th

Let a be the first term of GP with r being the common ratio.

∴ b = ar …(1)

c = ar² …(2)

Given,

(a + b + c) = xb

⇒ (a + ar + ar²) = x(ar)

⇒ a(1 + r + r²) = ar

⇒ (1 + r + r²) = xr

⇒ r² + (1 – x)r + 1 = 0

As r is a real number ⇒ Both solutions are real.

So discriminant of the given quadratic equation D ≥ 0

As, D ≥ 0

⇒ (1 – x)² – 4(1)(1) ≥ 0

⇒ x² – 2x – 3 ≥ 0

⇒ (x – 1)(x – 3) ≥ 0

∴ x < – 1 or x > 3 …proved