已知Sn为等差数列{an}的前n项之和,S9=18,Sn=256,a(n-4)=30（n〉9）,求n

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S9 = 18
(a1 + a9)*9 /2 = 18
a1 + a9 = 4
a1 + a1 + (9-1)*d = 4
a1 + 4d = 2
a1 = 2-4d
a(n-4)=30
a1 + (n-4-1)*d = 30
a1 + (n-5)d = 30
2-4d + (n-5)d = 30
(n - 9)d = 28
Sn = 256
(a1 + an)*n/2 = 256
[a1 + a1 + (n-1)d]*n = 512
[2a1 + (n-1)d]*n = 512
[4 - 8d + (n-1)d]*n = 512
[4 + (n-9)d]*n = 512
[4 + 28]*n = 512
32 n = 512
n = 16

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