G4.1 $x_{1}=2001$. When $n>1, x_{n}=\frac{n}{x_{n-1}}$. Given that $x_{1} x_{2} x_{3} \ldots x_{10}=a$, find the value of $a$.
$$
\begin{array}{l}
x_{2}=\frac{2}{x_{1}} \Rightarrow x_{1} x_{2}=2 \\
x_{4}=\frac{4}{x_{3}} \Rightarrow x_{3} x_{4}=4 \\
x_{6}=\frac{6}{x_{5}} \Rightarrow x_{5} x_{6}=6 \\
x_{8}=\frac{8}{x_{7}} \Rightarrow x_{7} x_{8}=8 \\
x_{10}=\frac{10}{x_{9}} \Rightarrow x_{9} x_{10}=10
\end{array}
$$

Multiply these equations gives $a=x_{1} x_{2} x_{3} \ldots x_{10}=2 \times 4 \times 6 \times 8 \times 10=32 \times 120=3840$