\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\) \(T_n = ar^n - 1\) ∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i) \(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii) = 2a = 3(1 - r) = 2a = 3 - 3r ∴ a = \(\frac{3 - 3r}{2}\) Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i) = \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\) = \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\) = 3(3r - 3r\(^2\)) = -4 = 9r - 9r\(^2\) = -4 = 9r\(^2\) - 9r - 4 = 0 = 9r\(^2\) - 12r + 3r - 4 = 0 = 3r(3r - 4) + 1(3r - 4) = 0 = (3r - 4)(3r + 1) = 0 ∴ r = \(\frac{4}{3} or - \frac{1}{3}\) For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1. ∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)