sn是an的前n项和,已知s636,sn324,sn-61

S6=(a1+a6)*6/2=362a1+5d=12Sn-S(n-6)=180即[a(n-5)+an]*6/2=180最后6项的和是6an-15d=1802an-5d=60相加2(a1+an)=72S

∵Sn=324，Sn-6=144，∴Sn-Sn-6=an-5+an-4+…+an=180又∵S6=a1+a2+…+a6=36，a1+an=a2+an-1=a6+an-5，∴6（a1+an）=36+18

因为da9|a3|=|a9|,a3>0,a9

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

公比q^3=a8/a5=16/(-2)=-8,则q=-2.a1=a5/q^4=-2/(-2)^4=-1/8.S6=a1(1-q^6)/(1-q)=-1/8*(1-(-2)^6)/(1+2)=-1/24

a4=a1q³a6=a3q³所以(a4+a6)/(a1+a3)=q³=(5/4)/10=1/8q=1/2a3=a1q²a1+a3=a1(1+q²)=1

a4/a1=q³a6/a3=q³则(a4+a6)/(a1+a3)=1/8=q³q=1/2a1+a3=a1+a1q²=5/4a1=10a1=8所以a5=a1q^4

S6=a1+a2+...+a6=(a3-2d)+(a3-d)+...+(a3+3d)=6a3+3d=7a3所以a3=3d,所以a1=d.a2+a7=a1+d+a1+6d=2a1+7d=9d=1,所以a

等差数列S3,S6-S3,S9-S6,S12-S9也成等差数列S3/S6=1/3,S6=3S3,S6-S3=2S3S9-S6=3S3,S9=6S3S12-S9=4S3,S12=10S3所以S6/S12

（Ⅰ）当q=1时，S3=3a1，S9=9a1，S6=6a1，∵2S9≠S3+S6，∴S3，S9，S6不成等差数列，与已知矛盾，∴q≠1．（2分）由2S9=S3+S6得：2•a1(1−q9)1−q＝a1

/>由于数列{an}为等比数列,则an=a1*q^(n-1),1）若q=1,S2=2a1,S6=6a1,S4=4a1S2,S6,S4成等差数列,则S2+S4=2S6,从而a1=0,不符合等比数列条件,

a2+a7=9所以2a1+7d=9(1)又S6=(a1+a6)*6/2=(2a1+5d)*3=7(a1+2d)所以a1=d(2)代入(1)中,所以a1=1=d所以an=a1+(n-1)d=1+(n-1

a6=3+5d(3+3+5d)*6*1/2=363*(6+5d)=366+5d=125d=6d=1.2an=3+1.2*(n-1)=1.8+1.2n

a1a5=a2a4=10S6=10105a6=36a6=11a3=5,公差为2an=2n-1

设数列{an}的首项为a,差值为m据题意有:Sn=na+n*(n-1)*m/2由S6=72,得a=12-5m/2.1又Sn=na+n*(n-1)*m/2=324将1代入分析当m=1,a=9.5,n=1

由题意,S9-S3=S6-S9而S9-S3=A4+...+A9S6-S9=-(A7+A8+A9)而(A4+A5+A6)+2(A7+A8+A9)=0A3(Q+Q²+Q²)+2A6(Q

1.A1q^3+A1q^6=2A1q^9.解之得q^3=12.当q=1时A2=A1A5=A1A8=A1所以A2+A5=2A8所以a2,a8,a5成等差数列

因为S(6)=6a(1)+6×5d/2=6a(1)+15dS(12)=12a(1)+66dS(18)=18a(1)+153d而S(12)-S(6)=6a(1)+51dS(18)-S(12)=6a(1)

an=a1+(n-1)d；Sn=(a1+an)*n/2S6=3(a1+a6);①S12-S6=3a1+6a12-3a6②S18-S12=3a1+9a18-6a12③②-①=6a12-6a6=6(a1+

设等比数列{an}首项a1,公比为qS3=a1+a2+a3S6-S3=a4+a5+a6=(a1+a2+a3)q^3=(q^3)S3S9-S6=a7+a8+a9=(a4+a5+a6)q^3=(q^3)&