设等差数列an共有3n项，它的前2n项之和是60，后2n项之和是100，求该数列中间n项的和

设前n项和为S1，中间n项和为S2，后n项和为S3.
S1=(A1+An)×n/2
中间n项可以看成以A(n+1)为首项A(2n)为末项的等差数列
S2=(A(n+1)+A(2n))×n/2=(A1+nd+An+nd)×n/2=(A1+An)×n/2+n^2×d=S1+n^2×d
同理
S3=(A(2n+1)+A(3n))×n/2=(A1+2nd+An+2nd)×n/2=(A1+An)×n/2+2n^2×d=S1+2n^2×d
S1+S2=2S1+n^2×d=60
S2+S3=2S1+3n^2×d=100
两式相减2n^2×d=40 n^2×d=20
2S1+20=60 S1=20
S2=20+20=40