Given that the sum of the first three terms of a sequence is 48, and the product of those three terms is 2496, what is the fourth term of the sequence?

I'm going to assume that you are referring to an arithmetic sequence, where the difference between consecutive terms of the sequence is constant.

Suppose that the first term is #a#, and the constant difference between each terms is #d#. Then the first three terms of the sequence are #a,a+d,a+2d#.

The sum of the first three terms is then

#a+(a+d)+(a+2d)=48#

#3a+3d=48#

#a+d=16#

which you correctly derived.

Now, the problem further gives that the product of the first three terms is

#a(a+d)(a+2d)=2496#

It may seem that the above equation is too difficult to solve. However, recall that you have already derived that #a+d=16#. Then,

#a(a+d)(a+d+d)=2496#

#16a(16+d)=2496#

#16a+ad=156#

Now, again using the fact that #a+d=16#, substitute #d=16-a# into the above equation:

#16a+a(16-a)=156#

#16a+16a-a^2=156#

#a^2-32a+156=0#

Factor the quadratic equation:

#(a-6)(a-26)=0#

Thus, we have #a=6,d=16-a=10# or #a=26,d=16-a=-10#.

So there are two possible sequences that satisfy the constraints defined in the problem:

#6,16,26,36,...#

#26,16,6,-4,...#

I'll leave it up to you to verify that these two series indeed work. Thus, you have the two possible answers of #36# and #-4#.