jmc

Because of symmetry, we may find all the possible values for $|a_n-a_{n-1}|$ and multiply by the number of times this value appears. Each occurs $5\cdot 8!$, because if you fix $a_n$ and $a_{n+1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ and $a_{n+1}$ can be. To find all possible values for $|a_n-a_{n-1}|$ we have to compute\begin{eqnarray} |1 - 10| + |1 - 9| + \ldots + |1 - 2|\\ + |2 - 10| + \ldots + |2 - 3| + |2 - 1|\\ + \ldots\\ + |10 - 9| \end{eqnarray} This is equivalent to $$2\sum _{k=1}^9\sum _{j=1}^{k}j=330$$ The total number of permutations is $10!$, so the average value is $\frac{330\cdot 8!\cdot 5}{10!}=\frac{55}{3}$, and $m+n=\boxed{58}$.