For all [integers](https://artofproblemsolving.com/wiki/index.php/Integer) $n$ greater than $1$, define $a_n = \frac{1}{\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1  \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$
By the [change of base](https://artofproblemsolving.com/wiki/index.php/Logarithms#Logarithmic_Properties) formula, $a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n$. Thus 
\begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ &= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\  &= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}