Given the geometric series ∑n=2∞7(3)n(−5)n a) Find r, the common ratio. b) Determine if the series converges or diverges. c) If it converges, find the limit.

Question

Given the geometric series ∑n=2∞7(3)n(−5)n  a) Find r, the common ratio.  b) Determine if the series converges or diverges.  c) If it converges, find the limit.

Velsenw · Accepted Answer

Given geometric series is ∑n=2∞7(3)n(−5)n (a) To find the common ratio: Consider the series, ∑n=2∞7(3)n(−5)n First term is 7×(−35)2=6325 Second term is 7×(−35)3=189125 Common ratio r=(7×(−35)3)(7×(−35)2)=−35 (b) To determine the series is convergent or divergent: Here common ratio is −35 |r|=|−35|=35&amp;lt;1, thus the series is convergent. (c) To find the limit: The limit is same as that of the sum of the series. Here ∑n=2∞7(3)n(−5)n is a geometric series with common ratio −35, also which is an infinite series. Sum of an infinite series is a1−r, where a is the first term and r is the common ratio. In the given series a=7×(−32)2,r=−35 Therefore the sum is (7×(−35)2)1−(−35)=(7×(−35)2)1+35=6340 Thus, the limit is 6340

Jeffrey Jordon · Accepted Answer

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Given the geometric series sum_{n=2}^inftyfrac{7(3)^n}{(-5)^n} a) Find r, the common ratio. b) Determine if the series converges or diverges. c) If it converges, find the limit.

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