If a , b and c be three distinct real numbers in G.P. and a + b + c = x b , then x cannot be:

Step-by-step solution

Given a ,   b ,   c are in G.P. Let common ratio of G.P. be r , then b = a r ,   c = a r 2 Given equation can be written as a + a r + a r 2 = x .   a .   r ⇒ x = r + 1 r + 1 As we know r > 0 , then r + 1 r ≥ 2 ⇒ x ≥ 3 r < 0 , then r + 1 r ≤ - 2 ⇒ x ≤ - 1 ⇒ x ∈ - ∞ , - 1 ∪ 3 ,   ∞ Hence, x   ≠ 2 .