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May 8th, 2015, 04:58 AM   #1
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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.
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May 8th, 2015, 05:16 AM   #2
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Originally Posted by matisolla View Post
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

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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.
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May 8th, 2015, 05:43 AM   #3
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Quote:
Originally Posted by Country Boy View Post
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.
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May 8th, 2015, 06:32 AM   #4
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3a + 3d = 24, a + d = 8, b = 8. 440/8 = 55, a = 5, c = 11.
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May 8th, 2015, 07:40 AM   #5
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I think you are assuming that $a$ and $c$ are both positive integers, Greg1313
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May 8th, 2015, 07:57 AM   #6
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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.
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May 8th, 2015, 08:00 AM   #7
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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.
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May 8th, 2015, 09:40 AM   #8
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Hello, matisolla!

Quote:



Let = the common difference.















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May 8th, 2015, 07:39 PM   #9
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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.
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May 9th, 2015, 07:03 PM   #10
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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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