已知等差数列{an}的前几项和sn=63,a4=9,a9=-1,则n=?

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已知等差数列{an}的前几项和sn=63,a4=9,a9=-1,则n=?
已知等差数列{an}的前几项和sn=63,a4=9,a9=-1,则n=?

已知等差数列{an}的前几项和sn=63,a4=9,a9=-1,则n=?
a4=a1+3d=9 (1) a9=a1+8d=-1 (2)
(2)-(1):5d=-10 d=-2 a1= -1+16=15
(a1+an)n=126
[15+15-2(n-1)]n=126
[30-2n+2]n=126
[32-2n]n=126
2n²-32n+126=0
n²-16n+63=0
(n-7)(n-9)=0
∴ n=7 or n=9

设公差为d
a9-a4=5d=-1-9=-10
d=-2
a1=a4-3d=9-(-2)×3=15
Sn=na1+n(n-1)d/2
15n+(-2)n(n-1)/2=63
整理，得
n²-16n+63=0
(n-7)(n-9)=0
n=7或n=9

A9=a4+5d=-1
d=-2
an=a4-2(n-4)=9-2n+8.=-2n+17
a1=15
Sn=(a1+an)n/2
63=n(15-2n+17)/2
n²-16n+63=0
n=7 or n=9

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