已知等差数列{an}中,Sn为{an}的前n项和,S3=15,a5=-1.求{an}的通项an与Sn
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a1+a3 = 2a2
所以 a1+a2+a3 = 3a2 = 15
a2 = 5
a5 - a2 = 3d = -1 -5 = -6
d = -2
a1 = a2 -d = 5 - (-2) = 7
所以an = a1 + (n-1)d = 7 + (-2)(n-1) = 9-2n
Sn = (a1+an)/2 * n = (7 + (9-2n))/2 * n = (8-n)*n = 8n - n²
所以 a1+a2+a3 = 3a2 = 15
a2 = 5
a5 - a2 = 3d = -1 -5 = -6
d = -2
a1 = a2 -d = 5 - (-2) = 7
所以an = a1 + (n-1)d = 7 + (-2)(n-1) = 9-2n
Sn = (a1+an)/2 * n = (7 + (9-2n))/2 * n = (8-n)*n = 8n - n²
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