a(1)=s(1)=p+q+r
a(n+1)=s(n+1)-s(n)=p(2n+1)+q=2p(n+1)-p+q,
a(2)=2p*2-p+q=3p+q,a(2)-a(1)=3p+q-(p+q+r)=2p-r.
a(3)-a(2)=2p
只有r=0时,才有a(2)-a(1)=a(3)-a(2),
此时,
a(n)=2np-p+q=p+q+(n-1)*2p,n=1,2,...
{a(n)}是首项为p+q,公差为2p的等差数列.