Question

等差数列{an}的前n项和为Sn,已知am-1+am+1-=0,S2m-1=38,则m=(　　)

(A)38(B)20(C)10(D)9

 

Answer 1

C

【解析】因为{an}是等差数列,所以am-1+am+1=2am,由am-1+am+1-=0,得2am-=0,所以am=2(am=0舍),又S2m-1=38,即=38,即(2m-1)×2=38,解得m=10,故选C.