1,a6+a7+a8=S8-S5=85-40=45
2,a_(n+1)-an=3n+2
an=(an-a_n-1)+((a_n-1)-a_(n-2))+...+(a2-a1)+a1
=(3n-1)+(3n-4)+..+5+2=（2+3n-1）*n/2=(3n+1)*n/2
3,因?yàn)锳、B、C共線,所以a1+a2007=1,
S2007=(a1+a2007)*2007/2=2007/2
4,a1+a3=2a2=-1,f(x-1)+f(x)=-1,即解得x=2或x=3,
所以a1=f(1)=0,或a1=f(2)=-1,對(duì)應(yīng)得a3=-1,或0
故公差為d=a2-a1=±0.5,所以an=a1+(n-1)d=(1-n)/2或an=(n-3)/2
5,第一個(gè)數(shù)列有y-x=3d；第二個(gè)數(shù)列有y-x=4d'
故a2-a1=d=(y-x)/3; b2-b1=d'=(y-x)/4,又xy
故原式=4/3
6,數(shù)列｛1/（an+1)}的公差d*4=[1/(1+a7)]-[1/(1+a3)]=1/2-1/3=1/6
故d=1/24,所以1/(1+a8)=d+1/(1+a7)=1/24+1/2=13/24,
所以a8=11/13
7,an=㏒2(n²+3)-㏒2(4)=log2[(n^2+3)/4
令an=log2(3),則（n^2+3）/4=3,又n>0,解得n=3,
所以為第三項(xiàng)