Question

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

Answer 1

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n<=7时an<0，当n>=8时an>0；
设Tn为{|an|}的前n项和
当n<=7时，Tn=－Sn
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

Answer 2

S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=3，a1=-20
∴an=a1+（n-1）d=3n-23

追问

后面咧...

Answer 3

S4=4a1+(4x3)/2d=-62
S6=6a1+(6x5)/2d=-75
d=-28，a1=57.5
∴an=a1+（n-1）d=57.5+（n-1）x(-28)=-28n+85.5
由-28n+85.5>0且n为正整数得n=1,2,3
所以,当n小于等于3时，︳a1︳+︳a2︳+︳a3︳+…+︳an|=na1+n(n-1)d=57.5n+n(n-1)x(-28)
当n大于3时，︳a1︳+︳a2︳+︳a3︳+…+︳an|=-sn+2s3=....
(后面不写了，应该能看懂，能知道后面的应该如何做了)