a1+a1+2d+a1+4d=3a1+6d=105,a1+2d=35
a1+d+a1+3d+a1+5d=3a1+9d=99,a1+3d=33
d=-2,a1=39
am<=0,a1+(m-1)d<=0
39-2m+2<=0
m>=41/2=21.5
m=21时,a21=39+20*-2=-1,a20=1
所以,S20=(39+1)*20/2=400,n=20
2、
a1+3d=1
(a1+a1+4d)*5/2=10,a1+2d=2
d=-1,a1=4
a5=4-4=0
S5=S4=10,取得最大值的n=4,n=5
3、
S17=(25+25+16d)*17/2=(25+25+8d)*9/2
50*17+16*17d=50*9+72d
200d=-400,d=-2
am<=0
25+(m-1)*(-2)<=0
25-2m+2<=0,m>=27/2=13.5
a13=25-24=1
S13=(25+1)*13/2=169