Let d be the common difference, then 

a=y-2d, x=y-d, y, z=y+d, b=y+2d

xyz = (y-d)y(y+d) = 15/2
         y(y²-d²) = 15/2
           y³-d²y = 15/2
         2y³-2d²y = 15
      2y³-2d²y-15 = 0

Since a and b are positive integers, a+b is a positive integer,
therefore a+b = y-2d+y+2d = 2y is a positive integer.
Let 2y = p, a positive integer.  Substitute y = p/2

2(p/2)³-2d²(p/2)-15 = 0
       2p³/8-d²p-15 = 0
        p³/4-d²p-15 = 0
         p³-4d²p-60 = 0    

Since p³ = 4d²p+60, p³ is a multiple of 4,
so is p.  4 is the only multiple of 4 which is a factor
of 60 and thus that can be a rational solutional to the cubic.  

We use synthetic division with p=4:

4 | 1  0    -4d²      -60
  |    4      16  64-16d²
    1  4  16-4d²   4-16d²

That is a solution if and only if 4-16d² = 0
                                   -16d² = -4
                                      d² = 1/4
                                       d = ±1/2

So we have p = 4, d = ±1/2

So y = p/2 = 4/2 = 2

a=y-2d, x=y-d, y, z=y+d, b=y+2d

Using d = 1/2
a=2-2(1/2)=1, x=2-(1/2)=3/2, y=2, z=2+(1/2)=5/2, b=3

So the positive integers are a=1 and b=3

Using d = -1/2
a=2-2(-1/2)=3, x=2-(-1/2)=5/2, y=2, z=2+(-1/2)=3/2, b=1

So the positive integers are a=3 and b=1

Either way the integers are 1 and 3.

Edwin