若等差数列{an}满足递推关系a(n+1)=-(an)+n,则a10=
问题描述：
若等差数列{an}满足递推关系a(n+1)=-(an)+n,则a10=
答
∵a(n+1)=-(an)+n∴an+a(n+1)=na(n+1)+a(n+2)=n+1a(n+2)-an=1∵an是等差数列∴d=1/2∴设a(n+1)=an+d∴an+a(n+1)=nan+an+d=nan=(n-d)/2an=(n-1/2)/2n=10a10=(10-1/2)/2=5-1/4=19/4