问题 选择题

已知等差数列{an}的前n项和为Sn,若S9=18,Sn=240,an-4=30,则n=(  )

A.18

B.17

C.16

D.15

答案

由等差数列的性质可得S9=

9(a1+a9)
2
=
9×2a5
2
=18,

解得a5=2,故a5+an-4=32,

而Sn=

n(a1+an)
2
=
n(a5+an-4)
2
=16n=240,解得n=15,

故选D

名词解释
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