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June 7th, 2017, 12:38 PM   #1
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Unhappy 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(?)

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June 7th, 2017, 01:35 PM   #2
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$a + (a+d) + (a+2d) = 24$

$a(a+d)(a+2d) = 440$

solve the system of equations for $a$ and the common difference of the arithmetic progression, $d$
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June 7th, 2017, 01:36 PM   #3
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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=(8-d)8(8+d)=440 => (8-d)(8+d)=55 => d=3
a, b, c = 5, 8, 11
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June 7th, 2017, 04:23 PM   #4
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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?
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June 7th, 2017, 08:10 PM   #5
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Quote:
Originally Posted by Asami View Post
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?
An arithmetic progression is a sequence of numbers that have a common difference between each consecutive pair of terms (that would be the $d$).

examples ...

1, 4, 7, 10, ... note $d=3$

14, 9, 4, -1, ... note $d=-5$

x, x-2a, x-4a, x-6a, ...
so, what is the common difference, $d$, of this progression?

Recommend you visit the link ...

arithmetic sequences & sums
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