I suppose
denotes the n-th term of some sequence, and we're given the 3rd and 5th terms
and
. On this information alone, it's impossible to determine the 100th term
because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.
To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.
• Assumption 1: the sequence is arithmetic (a.k.a. linear)
In this case, consecutive terms <u>d</u>iffer by a constant d, or

By this relation,

and by substitution,

We can continue in this fashion to get


and so on, down to writing the n-th term in terms of the first as

Now, with the given known values, we have


Eliminate
to solve for d :

Find the first term
:

Then the 100th term in the sequence is

• Assumption 2: the sequence is geometric
In this case, the <u>r</u>atio of consecutive terms is a constant r such that

We can solve for
in terms of
like we did in the arithmetic case.

and so on down to

Now,


Eliminate
and solve for r by dividing

Solve for
:

Then the 100th term is

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...