If the sum of four consecutive terms of a G.P. with real terms is 45 and the sum of their squares is 765 , then the sum of all possible values of the common ratio of this G.P. is

Step-by-step solution

Let the four consecutive terms of the G.P. be a, ar, ar^2, ar^3 . The sum of the terms is: S = a(1 + r + r^2 + r^3) = a(1+r)(1+r^2) = 45 The sum of their squares is: S₂ = a^2(1 + r^2 + r^4 + r^6) = a^2(1+r^2)(1+r^4) = 765 Consider the ratio S^2 S₂ : S^2 S₂ = a^2(1+r)^2(1+r^2)^2 a^2(1+r^2)(1+r^4) = (1+r)^2(1+r^2) 1+r^4 = 45^2 765 = 2025 765 = 45 17 Divide the numerator and the denominator of the algebraic fraction by r^2 : Numerator: (1+r)^2 r 1+r^2 r = (r + 1 r + 2 ) (r + 1 r ) Denominator: 1+r^4 r^2 = r^2 + 1 r^2