设等差数列｛an｝的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/12 06:22:21

S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=3,a1=-20
∴an=a1+（n-1）d=3n-23
由an>0即3n-23>0得 n>23/3
所以当n0；
设Tn为{|an|}的前n项和
当n=8时,Tn=|a1|+|a2|+|a3|+…+|a7|+|a8|+…+|an|
=-a1-a2-a3-…-a7+a8+…+an
=Sn-2S7
再代入计算就行了

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