1、∵在等差數(shù)列中
S4=4*a1+4*(4-1)*d/2=4a1+6d= -62
S6=6*a1+6*(6-1)*d/2=6a1+15d= -75
整理得：
2a1+3d= -31……(1)
2a1+5d= -25……(2)
由(2) - (1)得：
2d=6
d=3
將d=3代入(1)得：
2a1+3*3 = -31
a1= -20
∴an=a1+(n-1)*d
= -20+(n-1)*3
=3n-23
Sn=n*a1+n*(n-1)*d/2
=n*(-20)+n*(n-1)*3/2
=3/2*n^2-43/2*n
2、∵an=3n-23
∴當an>0時
3n-23 > 0
n > 23/3
即當n=7時,an0
∴a1,a2,a3,a4,a5,a6,a7都是負數(shù)
a8之后都是正數(shù)
∴|a1|+|a2|+…+|a24|
= -(a1+a2+…+a7)+(a8+a9+…+a24)
= -S7+(S24-S7)
=S24-2S7
=[3/2*24^2-43/2*24] - 2*[3/2*7^2-43/2*7]
=348 - (-154)
=502