答：p≠1,q=-1充分性Sn=p^n+q为等比数列,pq≠0S(n+1)=p^(n+1)+q,两式相减,A(n+1)=p^n*(p-1),由题意,当n=0也成立,A1=p+q=p-1,q=-1,An=p^(n-1)*(p-1),故p≠1,必要性p≠1,q=-1,pq≠Sn=p^n-1A1=p-1S(n+1)=p^(n+1)-1A(n+1...