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eliezer25

22.05.2021

Mathematics

The geometric series is given by:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

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 $n^{th}$ term of a geometric sequence is given by:

$a_n = ar^{n-1}$

There are 16 teams remaining in round 4 and 4 teams in round 6.

Thus, we can write:

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4}=\dfrac{ar^3}{ar^5}\\\\4=\dfrac{1}{r^2}\\r^2=\dfrac{1}{4}\\\\r = \dfrac{1}{2}$

Putting value of r in he equation we get:

$16=a(\dfrac{1}{2})^2\\\\a = 16\times 2^3\\a = 128$

Thus, the geometric sequence can be written as:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

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