May 8th, 2015, 04:58 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  a+b+c=24, abc=440. Arithmetic sequence
Assume a, b and c are consecutive terms of an arithmetic progression. (a smaller than b smaller than c). If a+b+c=24 and abc=440, find the values of a, b and c. 
May 8th, 2015, 05:16 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
Quote:
abc= a( 8 )(16 a)= 128a 8a^2= 440. Dividing through by 8, 16a a^2= 55 or a^2 16a+ 55= 0. That is easily factorable.  
May 8th, 2015, 05:43 AM  #3  
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Quote:
Lol I had gotten to a^2 16a+ 55= 0 with a different procedure but instead of writing 55 I wrote 56, so the answer was all weird haha silly mistake! Thanks for helping me notice. Last edited by skipjack; May 8th, 2015 at 06:24 PM.  
May 8th, 2015, 06:32 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
3a + 3d = 24, a + d = 8, b = 8. 440/8 = 55, a = 5, c = 11.

May 8th, 2015, 07:40 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
I think you are assuming that $a$ and $c$ are both positive integers, Greg1313 
May 8th, 2015, 07:57 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
No, a and c are positive integers. a + b + c = 24 implies b = 8 and 440/8 = 55, hence a must be 5 and c must be 11.

May 8th, 2015, 08:00 AM  #7 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
They must be. As $\displaystyle a < b < c$ and $\displaystyle abc = 440 > 0$, either $\displaystyle a < b < 0 < c$ or $\displaystyle 0 < a < b < c$. As $a$, $b$ and $c$ are in arithmetic progression and so $b = 8$, we must have $0 < a < b < c$, i.e. $a$ and $c$ are positive integers.
Last edited by skipjack; May 8th, 2015 at 06:17 PM. 
May 8th, 2015, 09:40 AM  #8  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, matisolla! Quote:
Let = the common difference.  
May 8th, 2015, 07:39 PM  #9 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 391 Thanks: 70 
Thank you, soroban. Ingenious! That's what I was looking forâ€”an algebraic solution that I can understand. I couldn't find one with my own efforts. Last edited by Timios; May 8th, 2015 at 07:42 PM. 
May 9th, 2015, 07:03 PM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Let the 3 terms be: ad , a , a+d sum = 3a = 24 ; a = 8 product: (8d)(8)(8+d) = 440 : d^2 = 9 : d = +3 or 3 so 5,8,11 or 11,8,5 Last edited by Denis; May 9th, 2015 at 07:06 PM. 

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arithmetic sequence a b c,in an arithmetic progression last 10 terms is 440.,a b c=24,solve 8a 440128a=0,a b c = 24 soal may,a b c=24. abc=440,,a b c=24 y abc=440,A=?B=16,C=24. ABC=8,B Knho c 24 horepr,if abc=24, a b c=11 what is a , b and c,a b c= 24 and abc=440 then a=,If a=8, b=3 and c=5 are integers, then value of a*b/c is A) 4 B) 2.8 C) 2.8 D) 3
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