设数列an是等差数列,其前n项和为s,若a6=2

1.已知{an}是等差数列,其首项是a1,公差为d,前n项和为Sn,由a11=11,s11=153得a1+10d=11且11a1+[11*(11-1)/2]d=153解得a1=185/11,d=-32

设{A(n)}的通项公式为：A(n)=2+d(n-1){B(n)}的通项公式为：B(n)=2×q^(n-1)则{A(n)}的前n项和为：S(n)=[A(1)+A(n)]n/2=[4+d(n-1)]n/

Sn=2an-2n则Sn+1=2an+1-2（n+1）an+1=Sn+1-Sn=2an+1-2an-2则an+1-2an=2所以{an+1-2an}是等差数列（2）an+1-2an=2则an+1+2=

an=Sn-Sn-1=n(a1+an)/2-(n-1)(a1+an-1)/22an=na1+nan-na1-nan-1+a1+an-1(n-2)an=(n-1)*(an-1)-a1(1)同理(n-1)

2Sn=an+an^22Sn-1=an-1+an-1^2两式相减：2an=an^2+an-(an-1+an-1^2)an^2-an-(an-1+an-1^2)=0(an-(an-1+1))(an+an

a9²=a15²a9²-a15²=0(a9-a15)(a9+a15)=0公差d不等于0所以a9+a15=0a1+8d+a1+14d=0a1+11d=0-----

2sn=(an)^2+an,2(sn+1)=（an+1）^2+(an+1)作差((sn+1)-(sn)=an+1)则((an+1)-an-1)((an+1)+an)=0因为数列{an}的各项都是正数所

解题过程如下.

题目中(am－an)/(m+n)是错的,应改为(am－an)/(m－n).必要性：an是公差为d的等差数列,则am＝a1+（m－1）d,an＝a1+（n－1）d,2S（m+n）＝2（m+n）a1+（m

∵a2=-6，a8=6∴a1+d=-6，a1+7d=6得a1=-8，d=2∴S4=S5故选B

证：第一种方法Sn+1=(n+1)[a1+a(n+1)]/2Sn=n(a1+an)/2Sn-1=(n-1)[a1+a(n-1)]/2a(n+1)=Sn+1-Sn=(n+1)[a1+a(n+1)]/2-

a3=a1+2d=6S3=a1+a2+a3=3a1+3d=12解得a1=2,d=2,故an=2n所以Sn=n(n+1)所以1/S1+1/S2+……+1/Sn=1/(1*2)+1/(2*3)+1/(3*

a2等于-6,a8等于6a1+d=-6a1+7d=6d=2,a1=-8S4=4a1+(4-1)4*2/2=-32+12=-20S5=5a1+(5-1)5*2/2=-40+20=-20S6=6a1+(6

Sn=a1n+n(n-1)d/2S4=4a1+6d=-62S6=6a1+15d=-75a1=-20,d=3an=a1+(n-1)d=3n-23当n＜8时,an＜0当n≥8时,an＞0|a1|+|a2|

a[n]等差数列所以：S[5]=5*a[3]=30,a[3]=6a[6]-a[3]=3d,2-6=3d,d=-4/3a[4]=6-4/3=14/3a[5]=6-2*4/3=10/3所以S[8]=4*(

（1）设首项和公差分别为a1，d由a3＝7S4＝24得a1+2d＝74a1+6d＝24所以a1＝3d＝2，则an=2n+1；（2）2Sp+q-（S2p+S2q）=2（p+q）2+4（p+q）-4p2-

因为Sn=3n^2+5nS（n-1）=3（n-1）^2+5(n-1)两式相减所以an=6n-3+5=6n+2所以an=8+6（n-1）,所以an是以8为第一项,公差为6的等差数列.

Sn=n(A1+An)/2设Bn=Sn/n=(A1+An)/2Bn-B(n-1)=(A1+An)/2-[A1+A(n-1)]/2=[An-A(n-1)]/2=d/2=常数∴｛Sn/n}是等差数列

T1=a1=1-a12a1=1a1=1/2a1a2...an=Tn=1-an(1)a1a2...a(n-1)=Tn-1=1-a(n-1)(2)(1)/(2)an=(1-an)/[1-a(n-1)]整理

a8=a2+(8-2)d所以:6=-6+6d得：d=2a1=a2-d=-6-2=-8Sn=na1+[n(n-1)d/2]=-8n+n(n-1)=n^2-9nS4=16-36=-20S5=25-45=-