已知等差数列中,S3=21,S6=24,求数列的前n项和Tn

S3=S8,3a1+3d=8a1+28da1=-5dS2=Sn2a1+d=na1+d[n(n-1)]/2-9=-5n+n(n-1)/2-18=-10n+n^-nn^-11n+18=0n1=2(舍去),

S2=q/4、S3=q^2/4、S4=q^3/4.由题意知,q^3/2=q/4+q^2/4,即2q^2-q-1=0,q=1或q=-1/2.1)q=1,则an=1/4,bn=2,1/[bnb(n+1)]

6(a1+a6)/2=s63(a1+a3)/2=s3because4s6=11s324(a1+a6)=33(a1+a3)24a1-33a1+24a6-33a3=obecausea6-a3=1thus-

题目为：a1+a2+a3=48a2*a2=a1*a4an=a1+(n-1)d求解a1、d即可带入公式可把条件转化为3a1+3d=48(a1+d)(a1+d)=a1(a1+3d)解出a1=16d=0或：

（1）设等差数列{an}的公差为d，∵前n项和Sn满足S3=0，S5=-5，∴3a1+3d＝05a1+10d＝−5，解得a1=1，d=-1．∴an=1-（n-1）=2-n．（2）1a2n−1a2n+1

等比数列求和公式为Sn=a1(1-q^n)/(1-q)s2=a1(1-q^2)/(1-q)=9s3=a1(1-q^3)/(1-q)=21解得a1=3q=2刚才题目没看清楚不好意思再问：没有其他可能了吗

A6+A8=2A7=26所以A7=13设公差为d,则S3=3A2=3(13-5d)=A5=13-2d解得d=2S10=7A7+S3=91+9=100

1、a3=6,S3=12所以a1+2d=6,3a1+3d=12所以a1=2,d=2所以{an}的通项公式为an=2+(n-1)*2=2n2、Sn=2n+n(n-1)=n²+n=n(n+1)1

S3=3*a1+3dS6=6*a1+15d得a1=9,d=-2an=-2n=11前5项为正Tn=a1+a2+a3+a4+a5-(a6+……+an)=（9+1）*5、2+（-1-2n+11）*(n-5)

由S3=a1+a2+a3=3a2.a2=-9又所以d=4a1=-13.Sn=na1+[（n-1)dn]/2=50解得n=10

直接来s3=3*a1+3d=21s8=8*a1+8*7/2*d=8*a1+28d=136解方程得d=4a1=3sn=a1*n+n*(n-1)/2*d=2n^2+n

设公比为q,则a3=a1q^2a5=a1q^4由题意得2(S5+a5)=S3+a3+S4+a4即2(a1+a2+a3+a4+a5+a5)=a1+a2+a3+a3+a1+a2+a3+a4+a4整理得4a

S3=S83a1+3*2d/2=8a1+8*7d/23a1+3d=8a1+28d5a1+25d=05(-25)+25d=0d=5an=a1+(n-1)d=-25+(n-1)5>0n-1>5n>6n=7

S3=9=3a2,a2=3S9=3=9a5,a5=1/3d=(a5-a2)/3=-8/9S12=6(a6+a7)=6(2a5+3d)=6(2/3-8/3)=-12

(1)S3=a1+a2+a3=3a1+3d=6s8=a1+a2+...+a8=8a1+28d=-4所以d=-1a1=3所以an=3+(n-1)(-1)=4-n（2）bn=(4-4+n)q=nqsn=b

在等差数列中：Sn=na1+n(n-1)d/2所以：S3=3a1+3*(3-1)d/2=39+3dS11=11a1+11*(11-1)d/2=143+55d所以：39+3d=143+55d所以：d=-

{a_n}的公差d'=-3/2.S_3-S_1=a_2+a_3=2a_1-9/2S_2-S_3=-a_3=3-a_1=2a_1-9/23a_1=15/2a_1=5/2d=3-a_1=1/2再问：我看不

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

设ak=a1+(k-1)d（设d是公差）a3=a1+2d,Sn=n*a1+n(n-1)/2所以S3=3a1+3d,因为bn=1/Sn且a3b3=1/2,推出a=d,所以Sn=d*n（n+1）/2因为S

设等差数列首项为a,公差为d:S3=3a+3dS6=6a+15dS12=12a+66d由题意：S3/S6=1/3即：(3a+3d)/(6a+15d)=1/3解得：a=2d∴S6/S12=(6a+15d