SOLUTION: Actually my question is about Arithmetic progression which I am finding difficult to work out so I hope you can help me with this question. This is the question, "Find three number

Question 399349: Actually my question is about Arithmetic progression which I am finding difficult to work out so I hope you can help me with this question. This is the question, "Find three numbers in an Arithmetic Progression whose sum is 21 and whose product is 280." The two equations I have worked out are a+d=7 and a^3-7a^2+49a-63=0. Are they correct? Well that was as far as I could go so I hope yuo can explain further. Thank you very much.
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
a + d =7 ==> d = 7 - a.
Now a(a+d)(a + 2d) = 280.
By substitution,
a(a +7 - a)(a + 14 - 2a) = 280
<==> 7a(14 - a) = 280
<==> a(14 -a) = 40
<==>
<==> (a - 4)(a - 10) = 0
==> a = 4, 10.
Two possible sequences arise, {4,7,10} and {10,7, 4}, which are essentially the same. Therefore the three numbers are 4,7, and 10.
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