Question

等差数列{a_n}中，a_n≠0,若m＞1,且a_m-1-+a_m+1=0,S_2m-1=38,则m的值为(    )

A.38                   B.20                   C.19                  D.10

Answer 1

解析：由题设=a_m-1+a_m+1=2a_m,而a_m≠0,∴a_m=2.

    又∵S_2m-1=38,

∴38===2(2m-1).

    解之得m=10,故本题应选D.

答案：D