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An infinite geometric series has first term $328$ and a sum of $2009$. What is its common ratio?

Guest Jan 27, 2015

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Best Answer

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We have

2009 = 328/(1 -r) multiply bth sides by (1 -r0

2009 (1 - r) = 328

2009 - 2009r = 328 rearrange

2009 - 328 = 2009r

1681 = 2009r divide both sides by 2009

1681 / 2009 = r = 41 / 49

CPhill Jan 27, 2015

1⁺⁰ Answers

+128707

+5

Best Answer

We have

2009 = 328/(1 -r) multiply bth sides by (1 -r0

2009 (1 - r) = 328

2009 - 2009r = 328 rearrange

2009 - 328 = 2009r

1681 = 2009r divide both sides by 2009

1681 / 2009 = r = 41 / 49

CPhill Jan 27, 2015

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