已知一个等差数列共有2n+1项,其中奇数项之和为290,偶数项之和为261,则第n+1项为( )
A.30
B.29
C.28
D.27
∵奇数项和S1=
=290(a1+a2n+1) (n+1) 2
∴a1+a2n+1=580 n+1
∵数列前2n+1项和S2=
=290+261=551(a1+a2n+1)(2n+1) 2
∴
=S1 S2
=(a1+a2n+1) (n+1) 2 (a1+a2n+1)(2n+1) 2
=2n+1 n+1 290 551
∴n=28
∴n+1=29
故选B