If a, b, c are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x < – 1 or x > 3.
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