解题思路:由题意可知Sm-1=0,由Sm-1=0得到首项与公差和m的关系,把要比较的两数作差后代入
a
1
=
(2−m)d
2
,因式分解后由已知得结论.
由am=Sm=a1+a2+…+am-1+am,得Sm-1=0,
∴(m−1)a1+
(m−1)(m−2)d
2=0,
∵m>1,∴a1=
(2−m)d
2.
Sn-an=na1+
n(n−1)d
2−[a1+(n−1)d]
=(n−1)a1+
n2d−nd−2nd+2d
2=(n−1)•
(2−m)d
2+
n2d−3nd+2d
2
=
2nd−2d−mnd+md+n2d−3nd+2d
2=
(n−1)(n−m)d
2.
∵m>1,n>m,d<0,
∴Sn-an<0,即Sn<an.
故选:B.
点评:
本题考点: 等差数列的性质;等差数列的前n项和;零向量.
考点点评: 此题考查等差数列的性质,考查了等差数列的前n项和,训练了作差法比较两个数的大小,是中档题.