已知数列an的首项为a,前n项和为sn,且点(sn/n,sn+1/n+1)在直线y=x+p上,p为常数

=>
S(n+1)/(n+1） - S(n)/n = p
S1/1 = a1 = a
=>
Sn/n = a+p*(n-1) = pn + a-p
=>
Sn = pnn + (a-p)n
=>
an = Sn - Sn-1 = p(2n-1) + (a-p) = 2pn +a-2p

S10最大
则
S9<=S10>=S11
=>
a10>=0
a11<=0
=>
20p + 10-2p>=0
22p + 10-2p<=0
=>
-5/9<=p<=-1/2

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
=90p+100
又S9=72p+90
S11=110p+110
且S10最大
则：S10＞=S9，S10＞=S11
可以算出P属于[-5/9，-1/2]

1
点(sn/n,sn+1/n+1)在直线y=x+p上,就有
sn+1/n+1=