A.S_k+                               B.S_k++

C.S_k+-                  D.S_k+

设数列a_n}为等差数列，其前n项和为S_n，已知a₁+a₃+a₈=99，a₅=31，若?k∈N^*，使得对于?n∈N^*，总有S_n≤S_k，则k=

1. A.
   19
2. B.
   20
3. C.
   21
4. D.
   35或36