If (2a+55),(a+b)and (a-b)are first 3 terms of a geometric progresion find a in terms of b and find the sum of first 6 terms in b each of two possible g.p
just crank it out. You know the ratio is constant, so
(a+b)/(2a+55) = (a-b)/(a+b)
(a+b)(a+b) = (a-b)(2a+55)
a^2 + 2ab + b^2 = 2a^2 - 2ab + 55a - 55b
a^2 + (55-4b)a - (b^2 + 55b) = 0
a = [(4b-55) ± √(20b^2-220b+3025))]/2
You can play around with that and find some solutions such as
a = -5, b = -10, r = -1/3
sequence = 45, -15, 5, ...
S6 = 45(1 - 1/3^6)/(1 + 1/3) = 910/27
There is even a fractional solution, with
a = -11/2, b = -33/2