等差数列{an}的前n项和为Sn,公差d<0,若存在正整数m(m>1)使am=Sm,则当n>m时,Sn与an的大小关系为

1个回答

  • 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.