等差数列的前n项和为sn,S6>s7>s5

1）.S6=6*A3+3*d=>d=2.=>A1=1=>An=A1+(n-1)*d=2*n-12).Bn=2^An+2*n=2^(2*n-1)+2*n=4^n/2+2*nTn=(4^1+4^2+4^3

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

（1）∵S6-S2=a3+a4+a5+a6=3a4=27∴a4=9∵a1=3∴d=2an=2n+1（2）Sn=n2+2n由已知得Sn•Sn+2=8（an+1+1）2∴n（n+2）2（n+4）=8（2n

a5+a6=S6-S4=2S10=10*(a5+a6)/2=5(a5+a6)=10

因为Sn=324,s(n-6)=144所以最后六项和=324-144=180=a(n-5)+a(n-4)+,+an又S6=36=a1+a2+,+a6两侧同时相加,有6(a1+an)=216a1+an=

因为S5＞S6所以a1+a2+a3+a4+a5>a1+a2+a3+a4+a5+a6所以a601.2a3=a1+a53a4=a1+a7+a42a3-3a4=d-a1-6d=-(a1+5d)=-a6>0(

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

因为S6,S9,S3成等差数列所以S6+S3=2S9,所以2a1(1+q+q^2)+a1q3(1+q+q^2)=2a(1+q+q^2+.q^8)解得q^3=-12S3*(S12-S6)=2a1(1+q

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

因为S3.S9.S6成等差数列2S9=S3+S62a1(1-q^8)/(1-q)=a1(1-q^2)/(1-q)+a1(1-q^5)/(1-q)2(1-q^8)=2-q^2-q^52q^8=q^2+q

由已知,可得S3=A1(1-q^3)/(1-q);S9=A1(1-q^9)/(1-q);S6=A1(1-q^6)/(1-q);S3,S9,S6成等差数列,所以S3+S6=2S9,化简,得q^3+q^6

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

S6=(a1+a6)*6/2=36得a1+a6=12记为1式Sn-S(n-6)=[a(n-5)+an]*6/2=324-144=180得a(n-5)+an=60记为2式又a6+a(n-5)=a1+an

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

首项为a1,比为qS6=S3+S3*q^3S9=S3+S3*q^3+S3*q^6S3,S9,S6成等差数列,那么S3+S9=2S6,即S3+S9-2S6=0S3+S3+S3*q^3+S3*q^6-2S

因为Sn=na1+[n(n-1)/2]d所以3=3a1+3d(1)24=6a1+15d(2)(2)-(1)*2解得d=2a1=-1因为an=a1+(n-1)d所以a9=-1+(9-1)*2=15

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

1）因为an=a*q；Sn=a*(1-q^n)/(1-q)；S3=a*(1-q^3)/(1-q)；S6=a*(1-q^6)/(1-q)；S9=a*(1-q^9)/(1-q)；2*S9=S3+S6；约去

因为a3=5所以A2+A4=A1+A5=10则a6=11所以a3=a1+2da6=a1+5d得a1=1d=2即an=2n-1（2）bn=2^((an+1)/2)将an=2n-1代入就得到bn=2^n为