A geometric series has a finite sum a1/(1-r) when the common ratio, r, satisfies the inequality -1 < r < 1. Otherwise, the series diverges.
So, for the geometric series in part (a), r = -2. Consequently, the
series is divergent. For the geometric series in part (b), r = 3. Therefore, the series diverges. In part (c), the series is geometric with r = 1/2. Since r is between -1 and 1, the series converges and has sum 60/(1-(1/2)) = 120.