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
答
由等差数列的性质可得S9=
=9(a1+a9) 2
=18,9×2a5
2
解得a5=2,故a5+an-4=32,
而Sn=
=n(a1+an) 2
=16n=240,解得n=15,n(a5+an−4) 2
故选D