Let: 
The common difference = d
The first term = a

Therefore:

The second term = a+d
The third term = a+d+d = a+2d

>>...The first three terms of an arithmetic series have a sum of 24...<<

           a + (a+d) + (a+2d) = 24
                   a+a+d+a+2d = 24
                        3a+3d = 24
                          a+d = 8
 
>>...The first three terms of an arithmetic series have...a product of 512...<<

                 a(a+d)(a+2d) = 512  
                 
So we have to solve the system:



Solve the first equation for d



Substitute in






Divide both sides by 8


Multiply through by -1

Factor:

Set each factor = 0

,     
      

There are two solutions for the first term, a,

Now since , taking the first solution:

When , 

so the first three terms are

, , 

Checking:  their sum = 13+8+3=24
           their product = (13)(8)(3)=312

So the fourth term is the third term plus (-5), or

fourth term = 3-5 = -2

-----------------------------

As before, since , taking the second solution:

When , 

so the first three terms are

, , 

Checking:  their sum = 3+8+13=24
           their product = (3)(8)(13)=312

So the fourth term is the third term plus 5, or

fourth term = 13+5 = 18

So there are two solutions for the fourth term: -2, and 18

Edwin