Question

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 can manage to show my work until
`16 =a+d`,
but I'm unsure about how to get further without guessing and checking.

Answer 1

`36` or `-4`

Explanation:

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`.