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 |
解得a5=2,故a5+an-4=32,
而Sn=
n(a1+an) |
2 |
n(a5+an-4) |
2 |
故选D
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 |
解得a5=2,故a5+an-4=32,
而Sn=
n(a1+an) |
2 |
n(a5+an-4) |
2 |
故选D