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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: 7,932 Thanks: 1127 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 26th, 2013, 12:21 AM  #4 
Member Joined: Aug 2013 Posts: 32 Thanks: 1  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: 20,474 Thanks: 2039 
Yes.


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