My Math Forum Geometric progression problem. Help

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 November 25th, 2013, 03:45 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Geometric progression problem. Help I solved this question until half. In a geometric progression, the 5th term and the sum of the 2nd and 3rd terms are 4/9 and 16 respectively. Given that r<0, find a( the first term and the common ratio.
 November 25th, 2013, 05:24 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Geometric progression problem. Help ar^4 = 4/9 ar + ar^2 = 16 a = 16/(r + r^2) 16r^4/(r + r^2) = 4/9 144r^3 - 4r - 4 = 0, r = 1/3 (this is the only real root; there is an error somewhere) a = 36.
 November 25th, 2013, 11:22 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 Google finds this.
 November 26th, 2013, 12:21 AM #4 Member   Joined: Aug 2013 Posts: 40 Thanks: 3 Re: Geometric progression problem. Help I think the question should state r<1, not r<0.
 November 27th, 2013, 02:08 AM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Re: Geometric progression problem. Help I think the question is definitely incorrect. If it state that r<1, is the answer r=1/3?
 November 27th, 2013, 05:47 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 Yes.

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