Answer:
![a_n = 128\bigg((1)/(2)\bigg)^(n-1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rgl0d3v5rhkzxn89m769ghwuyet1xqkxmt.png)
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
term of geometric sequence is given by:
![a_n = ar^(n-1)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ms8kio35bpqd8h4ztji6u795rry9u4kuo9.png)
![a_4 = 16 = ar^3\\a_6 = 4 = ar^5](https://img.qammunity.org/2021/formulas/mathematics/high-school/nw862kb9f0veyi5zi5t2304h1ppktdf04l.png)
Dividing the two equations, we get,
![(16)/(4) = (ar^3)/(ar^5)\\\\4}=(1)/(r^2)\\\\\Rightarrow r^2 = (1)/(4)\\\Rightarrow r = (1)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/do45wjxh475yvro2mm4hnzj9p42oo8y44d.png)
the first term can be calculated as:
![16=a((1)/(2))^3\\\\a = 16* 6\\a = 128](https://img.qammunity.org/2021/formulas/mathematics/high-school/rdz9md09u5jrx3jhibts6swqzm6bwoczev.png)
Thus, the required geometric sequence is
![a_n = 128\bigg((1)/(2)\bigg)^(n-1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rgl0d3v5rhkzxn89m769ghwuyet1xqkxmt.png)