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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: 20,929 Thanks: 2205 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: 20,929 Thanks: 2205 The rational root can be found with the help of the rational root theorem.

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### 8r³-27 solve me

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