My Math Forum a+b+c=24, abc=440. Arithmetic sequence

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 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
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Quote:
 Originally Posted by matisolla Assume a, b and c are consecutive terms of an arithmetic progression. (a smaller than b smaller than c).
So b- a= c- b and, from that, 2b= a+ c and c= 2b- a

Quote:
 If a+b+c=24 and abc=440, find the values of a, b and c.
a+ b+ c= b+ 2b= 3b= 24 so b= 8. Then c= 2b- a= 16- a.

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
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Quote:
 Originally Posted by Country Boy So b- a= c- b and, from that, 2b= a+ c and c= 2b- a a+ b+ c= b+ 2b= 3b= 24 so b= 8. Then c= 2b- a= 16- a. 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.

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
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Hello, matisolla!

Quote:
 $a,\,b,\,c\text{ are consecutive terms of an A.P. }\;(a\,<\,b\,<\,c)$ $\text{If }a\,+\,b\,+\,c\:=\:24\,\text{ and }\,abc\,=\,440,\text{ find the values of }a,\,b,\,c.$

Let $d$ = the common difference.

$\text{Then: }\:(a,\,b,\,c) \;=\;(b-d,\,b,\,b+d)$

$\text{Hence: }\:a\,+\,b\,+\,c\:=\:24 \;\;\;\Rightarrow\;\;\;(b-d)\,+\,b\,+\,(b+d) \:=\:24$

$\;\;\;\;\;3b \:=\:24 \;\;\;\Rightarrow\;\;\;b \,=\,8$

$abc \:=\:440 \;\;\;\Rightarrow\;\;\;(8-d)8(8+d) \:=\:440$

$\;\;\;\;\; 64\,-\,d^2 \:=\:55 \;\;\;\Rightarrow\;\;\;d^2 \:=\:9 \;\;\;\Rightarrow\;\;\;d\,=\,3$

$\text{Therefore: }\:(a,\,b,\,c) \;=\;(5,\,8,\,11)$

 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 for—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: a-d , a , a+d sum = 3a = 24 ; a = 8 product: (8-d)(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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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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