Answer:
The geometric series is given by:
[tex]a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}[/tex]
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 [tex]n^{th}[/tex] term of a geometric sequence is given by:
[tex]a_n = ar^{n-1}[/tex]
There are 16 teams remaining in round 4 and 4 teams in round 6.
Thus, we can write:
[tex]a_4 = 16 = ar^3\\a_6 = 4 = ar^5[/tex]
Dividing the two equations, we get,
[tex]\dfrac{16}{4}=\dfrac{ar^3}{ar^5}\\\\4=\dfrac{1}{r^2}\\r^2=\dfrac{1}{4}\\\\r = \dfrac{1}{2}[/tex]
Putting value of r in he equation we get:
[tex]16=a(\dfrac{1}{2})^2\\\\a = 16\times 2^3\\a = 128[/tex]
Thus, the geometric sequence can be written as:
[tex]a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}[/tex]
