Let x be the first number in this sequence of four. Then the next three are x + n, x + 2n, and x + 3n, because consecutive terms in an arithmetic progression differ by a fixed constant.
The sum of these four numbers is 28:
x + (x + n) + (x + 2n) + (x + 3n) = 4x + 6n = 28
or
2x + 3n = 14
The product of the second and third numbers is greater than the product of the first and last numbers by 18:
(x + n)(x + 2n) = x(x + 3n) + 18
Expanding everything on both sides gives
x² + 3nx + 2n² = x² + 3nx + 18
and terms cancel so we end up with
2n² = 18 ==> n² = 9 ==> n = 3 or n = -3
If n = 3, then
4x + 6*3 = 28 ==> 4x = 10 ==> x = 5/2
If n = -3, then
4x + 6*(-3) = 28 ==> 4x = 46 ==> x = 23/2
So we have two possible solutions, but the only difference between them is the order. If n = 3, then the four numbers are 5/2, 11/2, 17/2, and 23/2. If n = -3, then the four numbers are, again, 23/2, 17/2, 11/2, and 5/2.
