Answer:
The geometric series is given by:

Step-by-step explanation:
We are given the following in the question:
The numbers of teams remaining in each round follows a geometric sequence.
Let a be the first term and r be the common ratio.
The
term of a geometric sequence is given by:

There are 16 teams remaining in round 4 and 4 teams in round 6.
Thus, we can write:

Dividing the two equations, we get,

Putting value of r in he equation we get:

Thus, the geometric sequence can be written as:
