Answer:
[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 the of the geometric sequence and r be the common ration.
The [tex]n^{th}[/tex] term of geometric sequence is given by:
[tex]a_n = ar^{n-1}[/tex]
[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}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}[/tex]
the first term can be calculated as:
[tex]16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128[/tex]
Thus, the required geometric sequence is
[tex]a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}[/tex]
