Question

等差数列{a_k}共有2n+1项（n∈N^*），其中所有奇数项之和为310，所有偶数项之和为300，则n=______．

Answer 1

∵奇数项和S₁=

(a1+a2n+1) (n+1)

2

=310
∴a₁+a_2n+1=

620

n+1

∵数列前2n+1项和S₂=

(a1+a2n+1)(2n+1)

2

=300+310=610
∴

S1

S2

=

(a1+a2n+1) (n+1) 2

(a1+a2n+1)(2n+1) 2

=

2n+1

n+1

=

310

610

∴n=30
故答案为：30