A geometric series has first term (11x−3), Second term (5x+3) and third term (3x−3) . (a) Find the two possible values of x  .For each of your values of x ,(b) find the corresponding value of the common ratio of the series.Given that the series is convergent,(c ) find the sum to infinity of the series.

(a) In a geometric series, the ratio between any term and the previous term is constant. Therefore, we can set up the following equations:

(5x+3)/(11x-3) = (3x-3)/(5x+3)

Cross-multiplying gives:

(5x+3)(3x-3) = (11x-3)(5x+3)

Expanding and simplifying gives:

15x^2 - 9x + 9x - 9 = 55x^2 + 33x - 15x - 9

This simplifies to:

40x^2 + 39x = 0

Factoring out x gives:

x(40x + 39) = 0

Setting each factor equal to zero gives the solutions x = 0 and x = -39/40.

(b) The common ratio for x = 0 is (50+3)/(110-3) = -1. The common ratio for x = -39/40 is (5*(-39/40)+3)/(11*(-39/40)-3) = -1.

(c) The sum to infinity of a geometric series is given by the formula S = a / (1 - r), where a is the first term and r is the common ratio. For x = 0, the first term is -3 and the common ratio is -1, so the sum to infinity is -3 / (1 - (-1)) = -3 / 2 = -1.5. For x = -39/40, the first term is (11*(-39/40)-3) = -15.15 and the common ratio is -1, so the sum to infinity is -15.15 / (1 - (-1)) = -15.15 / 2 = -7.575.