Let a be the first term and d be the common difference.

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

So 

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

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



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-d² = 168
         -d² = -1
          d² = 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