No, there does not. Let $a_n=a+nd$. $a_n+...+a_{n+9}=10(a+nd)+45d$, while $a_n...a_{n+9}=(a+nd)...(a+nd+9d)$. Observing that $\gcd (a+nd+md, 2(a+nd)+9d) \leq$ $ \gcd (2a+2nd+2md, 2a+2nd+9d) = \gcd (2a+2nd+9d, (2m-9)d) \leq (2m-9)d$, where $0 \leq m \leq 9$, it is clear that divisibility cannot occur once $n$ exceeds $(10!)d^{10}$.