An algebra problem by أحمد الحلاق

$\begin{aligned} A & = \sum_{n=2}^{2001} \frac 1{\log_{a_n} 10} = \sum_{n=2}^{2001} \frac 1{\frac {\log_{10} 10}{\log_{10} a_n}} = \sum_{n=2}^{2001} \log_{10} a_n \\ & = \sum_{n=2}^{2001} \log_{10} \left(\log_n (n+1) \right) \\ & = \sum_{n=2}^{2001} \log_{10} \left(\frac {\log_{10} (n+1)}{\log_{10} n} \right) \\ & = \log_{10} \left(\prod_{n=2}^{2001} \frac {\log_{10} (n+1)}{\log_{10} n} \right) \\ & = \log_{10} \left(\frac {\log_{10} 2002}{\log_{10} 2} \right) \end{aligned}$

Therefore, $\begin{aligned} 10^A & = \frac {\log_{10} 2002}{\log_{10} 2} \approx \boxed{10.967} \end{aligned}$