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
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