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June 7th, 2017, 12:38 PM  #1 
Newbie Joined: Dec 2016 From: Argentina Posts: 5 Thanks: 0  Arithmetic Progression exercise
Hi guys, I have a very important exam very soon and this is one of the examples. No idea how to do this one, can you help me? Assume that a, b and c are consecutive terms of an arithmetic progression (a<b<c). If a+b+c=24 and abc=440 then a(?), b(?) and c(?) Thanks 
June 7th, 2017, 01:35 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 
$a + (a+d) + (a+2d) = 24$ $a(a+d)(a+2d) = 440$ solve the system of equations for $a$ and the common difference of the arithmetic progression, $d$ 
June 7th, 2017, 01:36 PM  #3 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
b=a+d, c=a+2d where d is a difference of the arithmetic progression. a+b+c=3a+3d=24 => a+d=b=8 abc=(8d)8(8+d)=440 => (8d)(8+d)=55 => d=3 a, b, c = 5, 8, 11 
June 7th, 2017, 04:23 PM  #4 
Newbie Joined: Dec 2016 From: Argentina Posts: 5 Thanks: 0 
Thank you for the help guys! But the truth is that I don't understand what you did, what is that D there doing what?

June 7th, 2017, 08:10 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387  Quote:
examples ... 1, 4, 7, 10, ... note $d=3$ 14, 9, 4, 1, ... note $d=5$ x, x2a, x4a, x6a, ... so, what is the common difference, $d$, of this progression? Recommend you visit the link ... arithmetic sequences & sums  

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