等差数列{an}中, S10＝100, S100＝10, 求 S110

因为等差，假设a1=a,公差d，则有an=a+（n-1）d

s10=(a1+a10)*10/2
=(2a+9d)*5
=10a+45d=100

s100=(a1+a100)*100/2
=(2a+99d)*50
=100a+4950d=10

s110=(a1+a110)*110/2
=（2a+109d)*55
=110a+5995d

s100-s10*10
=4500d=-990
d=-0.22

s110=s10+s100+1000d=100+10-220=-110

等于-110,设首项为a1,公差为d,则Sn=na1+n(n-1)/2,由已知得 :S10=10a1+10*9/2*d=100,S100=100a1+100*99/2*d=10;解得d=-11/50.a1=1099/100 ,所以可以求出通项公式了啊,再将n=110代入就可以得到-110

假设d为等差数列的公差
S10 = (a1 + a1 + d * 9) * 5 = (2a1 + 9d) * 5 = 100
S100 = (a1 + a1 + d * 99) * 50 = (2a1 + 99d) * 50 = 10

解得：
a1 = 10.99
d = -0.22

所以：
S110 = (2a1 + 109d) * 55 = -110

s10=100 s100=10 Sn=110-n
S110=110-110=0

公差-1
a1=110
a110=0