Given:
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence
The general form of the geometric sequence is as follows:
[tex]a_n=a_1\cdot r^{n-1^{}}[/tex]
Given the following conditions:
1) There are 32 teams remaining in round 4
So,
[tex]\begin{gathered} 32=a_1\cdot r^{4-1} \\ 32=a_1\cdot r^3\rightarrow(1) \end{gathered}[/tex]
2) there are 8 teams in round 6
So,
[tex]\begin{gathered} 8=a_1\cdot r^{6-1} \\ 8=a_1\cdot r^5\rightarrow(2) \end{gathered}[/tex]
So, we will solve the equations (1) and (2) to find a1, and (r)
Divide the equation (2) by (1) to eliminate a1, then solve for (r)
[tex]\begin{gathered} \frac{8}{32}=\frac{a_1\cdot r^5}{a_1\cdot r^3} \\ \\ \frac{1}{4}=r^{^{5-3}}\rightarrow r^2=\frac{1}{4} \\ \\ r=\sqrt[]{\frac{1}{4}}=\frac{1}{2} \end{gathered}[/tex]
Substitute with (r) into equation (1) to find a1
[tex]\begin{gathered} 32=a_1\cdot(\frac{1}{2})^3 \\ 32=a_1\cdot\frac{1}{8} \\ a_1=32\cdot8=256 \end{gathered}[/tex]
So, the answer will be:
[tex]a_n=256\cdot(\frac{1}{2})^{n-1}[/tex]