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March 4th, 2016, 12:12 PM  #1 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0  Geometric sequence
This problem keeps bothering me: The sum of the first three terms of a geometric sequence is 23 3/4 (23.75) and the sum of the first four terms is 40 5/8. Find the first term and the common ratio. So I found that the fourth term is 16 7/8 (16.875). And then the sum of the first three terms of this GP would be: 16.875/r^3+16.875/r^2+16.875/r=23.75 I know that the common ratio is 1.5. But I am stuck here; how should I solve this equation of the third degree? And I do not think that I should approach this problem in this particular way. Is there any other way that makes more sense? Thank you. Last edited by skipjack; March 4th, 2016 at 12:33 PM. 
March 4th, 2016, 01:20 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,476 Thanks: 2039 
I think your way is best. Multiplying your equation by 8r³/5 gives 27 + 27r + 27r² = 38r³, so 38r³  27r²  27r  27 = 0, i.e., (2r  3)(19r² + 15r + 9) = 0. Hence r = 3/2 or 3(5 ± $\small\sqrt{51}i$)/38. 
March 4th, 2016, 11:30 PM  #3 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0 
Thank you for the response. Can you explain me this part? 38r³  27r²  27r  27 = 0, i.e., (2r  3)(19r² + 15r + 9) = 0. How we get the second part? Last edited by skipjack; March 5th, 2016 at 03:55 AM. 
March 5th, 2016, 03:53 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,476 Thanks: 2039 
The rational root can be found with the help of the rational root theorem.


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