smc

Note that $\frac{1}{{\log }_{a}b}={\log }_{b}a$. Thus $a_n={\log }_{2002}n$. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: $${\log }_{2002}\left(\frac{2\ast 3\ast 4\ast 5}{10\ast 11\ast 12\ast 13\ast 14}=\frac{1}{11\ast 13\ast 14}=\frac{1}{2002}\right)=\boxed{-1}$$