已知S12=8,S20=460,求S28

a1+a2+.+a20=170(a1+a20)=(a2+a19)=.=170/10=17a6+a9+a12+a15=a6+a15+a9+a12=17+17=34

等差数列{an}，S10=310，S20=1220，S10，S20-S10，S30-S20成等差数列，设S30=x，则2（1220-310）=310+（x-1220），解得x=2730．故答案为：27

由S10=100,S20-10代入等差求和公式Sn=nA1+[n(n-1)D]/2中可以得到10A1+[10(10-1)D]/2=100.公式120A1+[20(20-1)D]/2=10.公式2可求得

解题思路：数列································解题过程：

设数列｛an｝公差为d.S4=a1+a2+a3+a4=6S8-S4=a5+a6+a7+a8S12-S8=a9+a10+a11+a12易知a1+a2+a3+a4,a5+a6+a7+a8,a9+a10+a

S12>0,S1307d+24>0d>-24/7S13=(a1+a1+12d)*13/2=(2a1+12d)*13/2=13(a1+6d)=13(a1+2d+4d)=13(a3+4d)=13(12+4

S12=12a1+12×11d/2=12(a3-2d)+66d=12a3+42d=42d+12×12=42d+144>0d>-24/7S13=13a1+13×12d/2=13(a3-2d)+78d=1

等差数列求和公式：Sn=n*a1+n*（n-1)*d/2S12=12*a1+12*11*d/2=12*a1+66d>0得a1+5.5d>0S13=13*a1+13*12*s/2=13*a1+78d

a4+a17=a1+a20=8所以S20=(a1+a20)/2*20=80

a1+a3=2a2=6a2=3a2+a4=2a3=8a3=4d=a3-a2=4-3=1a1=a2-d=3-1=2a20=a1+19d=2+19=21S20=20a1+20×19×d/2=20×2+20

S12=a1+a12=(a1+a12)*6=[a1+a1+(12-1)*d]*6=(2a1+11d)*68S4=8*[(a1+a4)*2]=8*{[a1+a1+(4-1)*d]*2}=8*[(2a1+

等差数列{an}s4,s8-s4,s12-s8也成等差数列2(s8-s4)=s4+s12-s82s8-2s4=s4+s12-s8s12=3s8-3s4s12=3(s8-s4)s12=3*(4-8)s1

等差数列S4,S8-S4,S12-S8也为等差,根据等差中项等于两边项之和的二倍得S12=12

an=a1+(n-1)da4+a6=2a5=-4,a5=-2a3=a5-2d,a7=a5+2d(-2-2d)(-2+2d)=-12,d>0d=2a5=a1+4d=a1+8=-2a1=-10s20=20

S8=48,S12=168S12-S8=120即A9+A10+A11+A12=120所以有：A4+A5=48/4=12A10+A11=120/2=60所以公差d=(60-12)/(6+6)=4所以A4

∵等差数列{an}，s10=S20，设s10=S20=a，S30=b，∴a，0，b-a成等差数列，∴0=a+b-a，解得b=0．故答案为：0．

A11~A20=(S20-S10)=90(A11-A1)=(11-1)d~(A20-A10)=(20-10)d90-S10+100d=080+100d=0d=-0.8S10=10*A0+(1+10)d

S3/S6=[(a1+a3)*3/2]/[(a1+a6)*6/2]=1/3(9/2)(a1+a3)=3(a1+a6)3(a1+a1+2d)=2(a1+a1+5d)6a1+6d=4a1+10da1=2d

S4=S12(a1+a4)*4/2=(a1+a12)*12/22a1+2a4=6a1+6a12a1+a1+3d=3a1+3a1+33d33d-3d=a1+a1-3a1-3a130d=-4a1=-60d