解:由于点(sn/n,sn+1/n+1)在直线y=x-p上
则有:
S(n+1)/(n+1)=Sn/n-p
则:
S(n+1)/(n+1)-Sn/n=-p
则：{Sn/n}为公差为-p的等差数列
则:Sn/n=S1/1-p(n-1)
=a1-p(n-1)
则:Sn=a1n-p(n-1)n
=-pn^2+(a1+p)n
=-pn^2+(a+p)n
则：n>=2时,
an=Sn-S(n-1)
=[-pn^2+(a+p)n]-[-p(n-1)^2+(a+p)(n-1)]
=-2pn+(a+2p)
又a1=-2p*1+a+2p=a
则：
an=-2pn+(a+2p)（n属于N*）
(2)
S10=10(a1+a10)/2, a10=-20p+a+2p=-18p+a
=-90p+10
又S9=-72p+9
S11=-110p+11
且S10最大
则：S10＞=S9,S10＞=S11
-90p+10>=-72p+9
1>=18p
p=-110p+11
20p>=1
p>=1/20
故p的范围是[1/18,1/20]