等差数列已知sn求n
发布网友
发布时间：2023-04-23 10:42
共1个回答
热心网友
时间：2023-04-25 18:44
(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
