SOLUTION: The sum of three numbers in A.P is 39 and their product is 2184. Find the three numbers? Please solve this step-by-steps for me, as this will enable me understand how come about th

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Question 977005: The sum of three numbers in A.P is 39 and their product is 2184. Find the three numbers? Please solve this step-by-steps for me, as this will enable me understand how come about the result. Thanks.
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Let a be the first term and d be the common difference.

Then the three terms are a, a+d, and a+2d.

The sum of three numbers in A.P is 39... 
So 

a+(a+d)+(a+2d) = 39
    a+a+d+a+2d = 39
         3a+3d = 39    <-- divide thru by 3
           a+d = 13

...and their product is 2184.
So 

a(a+d)(a+2d) = 2184      <-- replace (a+d) by 13

a(13)(a+2d) = 2184     

Divide through by 13

a(a+2d) = 168

solve the system

system%28a%2Bd+=+13%2Ca%28a%2B2d%29+=+168%29

Solve a+d = 13 for a

a = 13-d

Substitute in

a(a+2d) = 168

(13-d)(13-d+2d) = 168

(13-d)(13+d) = 168
      169-dČ = 168
         -dČ = -1
          dČ = 1
           d = ±1

a = 13-d.  Using  d = 1,
a = 13-1.
a = 12

So the three numbers in A.P are 

a, a+d, and a+2d which become

12, 12+1, 12+2(1) or
12, 13, 12+2 or
12, 13, 14 

 = 13-d.  Using  d = -1,
a = 13-(-1).
a = 13+1
a = 14

So the three numbers in A.P are 

a, a+d, and a+2d which become

14, 14+(-1), 14+2(-1) or
12, 14-1, 14-2 or
14, 13, 12

So as it turns out, there are not really two solutions.  It's
just a matter of whether they are 12,13,14 or 14,13,12.

Edwin