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

问题描述:

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

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