If the sum of four positive consecutive terms of a G.P. is 112 and the sum of their reciprocals is 28 , then the sum of all possible values of the common ratio of such a G.P. is

Step-by-step solution

Let the four positive 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) = 112 The sum of their reciprocals is: S' = 1 a + 1 ar + 1 ar^2 + 1 ar^3 = 1 a ( r^3 + r^2 + r + 1 r^3 ) = 28 Dividing S by S' gives: S S' = a^2 r^3 = 112 28 = 4 Since the terms are positive, a and r are positive, yielding: a = 2 r^ 3/2 Substituting this into the sum equation: 2 r^ 3/2 (1 + r + r^2 + r^3) = 112 r^ -3/2 + r^ -1/2 + r^ 1/2 + r^ 3/2 = 56 Let t = r^ 1/2 + r^ -1/2 . Cubing both si