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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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
![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](/uploads/image/z/8391761-17-1.jpg?t=%E5%B7%B2%E7%9F%A5Sn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E4%B9%8B%E5%92%8C%2CS9%3D18%2CSn%3D256%2Ca%28n-4%29%3D30%EF%BC%88n%E3%80%899%EF%BC%89%2C%E6%B1%82n)
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
(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
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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