First, note that the maximal number of initial primes is bounded above by the smallest prime not dividing $A_{11}$, with equality possible only if $p$ is this prime. For, if $q$ is the smallest prime not dividing $A_{11}$, then the first $q$ terms of the arithmetic sequence determine a complete residue class modulo $q$, and the multiple of $q$ is nonprime unless it equals $q$. If $q<A_{11}$, then $q$ must appear first in the sequence, and thus divide the $(q+1)$ st term. If $q>A_{11}$, then $A_{11}=2$ and $q=3$ by Bertrand's postulate, so $q$ must appear first by inspection. Now since $A_{11}=30$, the bound is 7. In fact, $7,37,67,97,127$, and 157 are prime, but 187 is not. Then on the one hand, our bound of seven initial primes is not realizable. On the other hand, this implies an upper bound of six, and this bound is achieved by $p=7$. Smaller primes $p$ yield only one initial prime, so 7 is the answer.