Mark M. answered • 06/12/15

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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.

**A geometric series has a finite sum a**

_{1}/(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.**