Question

高二等差数列题目！！！！！！！！！！！！急

在等差数列{AN}中

已知S8=48 S12=168 求A1和D1

Answer 1

S8=48，S12=168
S12 - S8 = 120即A9 + A10 + A11 + A12 = 120

所以有：
A4 + A5 = 48/4 = 12
A10 + A11 = 120/2 = 60

所以公差d = (60 - 12)/(6 + 6) = 4
所以A4 = (12 - 4)/2 = 4
所以A1 = 4 - 4*(4 - 1) = -8