设等差数列前n项和为sn,已知A2=6,6A1 A3=30

1.S12=12*a1+12*(12-1)*d/2>0S13=13*a1+13*(13-1)*d/20且3+d0S12=(a1+a13)*13/2

等差数列前n项和Sn=na1+n*（n-1）*d/2n=6时S6=6a1+6*5*d/2S6=6a1+15d36=6a1+15da1=6-(5/2)dSn=na1+n*(n-1)*d/2=324将a1

（1）设等差数列{an}的公差为d，则Sn=na1+12n（n-1）d，∵S7=7，S15=75，∴7a1+21d=715a1+105d=75-----------------------------

s3=a1+a2+a3即9=a1+a2+5所以a1+a2=4因为a1+a3=2*a2所以合解得a1=1,a2=3,a3=5

S5=5(a1+a5)/2=5a3=25,则a3=5,d=a3-a2=2数列{an}是以a1=1为首项,d=2为公差的等差数列则通项式为an=a1+(n-1)d=2n-1bn=2（an）次方＋1则{b

1.s2/s1=c+1s2=c+1a2=cs3/s2=(2+c)/2s3=(2+c)(c+1)/2a3=c(c+1)/22a2=a1+a32c=1+c(c+1)/2c^2-3c+2=0c=1或22.c

提示A1=A3-2d=12-2dS12=12A1+66d=12(12-2d)+66d=144+42d>0,∴d>-144/42=-24/7.S13=13A1+78d=13(12-2D)+78d=156

s12=12(a1+a12)>0s13=13*(a1+a13)0,a3+a10>0,2a3+7d>0,24+7d>0a1+a13

（1）S12=12a1+12×(12-1)／2•d＞0,S13=13a1+13×(13-1)／2•d＜02a1+11d＞0①a1+6d＜0②a3=12,得a1=12-2d③,将

依题设有S12=12a1+12×11d/2>0S13=13a1+13×12d/20a1+6d03+d

(1)因为S12=12*(a1+a12)/2＞0,S13=13*(a1+a13)/2=13*a7＜0所以a1+a12＞0,a7＜0又a3=12所以(a3-2d)+(a3+9d)=24+7d＞0,a3+

通项公式为an=d(n-3)+12a1=12-2d,a12=9d+12,a13=10d+12s12=(a1+a12)×12/2>0s13=(a1+a13)×13/2

S12=12*a1+12*(12-1)*d/2>0S13=13*a1+13*(13-1)*d/20且3+d

（Ⅰ）∵等差数列{an}中，a3=5，S3=9，∴a1+2d＝53a1+3×22d＝9，解得a1=1，d=2，（Ⅱ）∵a1=1，d=2，∴Sn＝n+n(n−1)2×2=n2=100，∴n=10．

S8>0,S909*(a1+a9)/20a4-3d+a4+5d-242d

S3=S12∴S12-S3=0∴a4+a5+.+a12=0∴(a4+a12)*9/2=0∴a4+a12=0∴a8+a8=a4+a12=0∴a8=0∵d

A1=A3-2d=12-2dS12=12A1+66d=12(12-2d)+66d=144+42d>0,∴d>-144/42=-24/7.S13=13A1+78d=13(12-2D)+78d=156+5

设公差为dS12=(a3+a10)*6=(2a3+7d)*6=(24+7d)*6>0S13=a7*13=(a3+4d)*13=(12+4d)*130且12+4d

因为a1+a11=a3+a9所以S11=(a1+a11)*11/2=(a3+a9)*11/2=（24+a9）*11/2=0所以a9=-24所以d=(a9-a3)/6=-8a1=a3-2d=24+16=

等差数列an=p*（n-1)+a1,sn=（a1+an）*n/2=n*a1+p*(n-1)*n/2a3=2*p+a1=24s11=11*a1+55*p=0得a1=40,p=-8(1)an=-8n+48