设等差数列an的前n项和公式

依题意得：S4＝4a1+4×32d＝−62S6＝6a1+6×52d＝−75解得：d=3，a1=-20∴an=a1+（n-1）d=3n-23．

等差数列{an}的首项为a1,公差为ds3=9a1+a2+a3=9a1+a1+d+a1+2d=93a1+3d=9a1+d=3a2=3a9=17a2+7d=177d=14d=2a1=1an=1+(n-1

设公差da2=a1+da4=a1+3da10=a1+9dS10=(a1+a10)*10/2=5(2a1+9d)=1102a1+9d=22a1.a2.a4为等比数列a2*a2=a1*a4(a1+d)^2

等差数列求和公式：sn=(a1+an)*n/2an=3n-1a1=3*1-1=2a6=3*6-1=17s6=(2+17)*6/2=57

等差数列有公式啊根据公式直接带入就行了啊联立方程Sn=(a1+an)/2*nan=a3+(n-3)d

s3=9得出a2=3代入a5+a13=34,得d=2所以a=2n+1第2个问题不完整

a1=s1=3设an=3+（n-1）d,则a2=3+d,a3=3+2ds3=9+3d=15所以d=2故an=2n+1,sn=（n+2）n

【答案】an=2n+1【解析】s5-s2=a3+a4+a5=3·a4=27所以,a4=9,d=（9-3）÷3=2an=3+（n-1）·2=2n+1二十年教学经验,专业值得信赖!敬请及时采纳,在右上角点

a4=s4-a3=16-5=11;等差：a4-a3=11-5=6那么,a1=a3-2*6=5-12=-7通项公式：an=a1+6n=-7+6n

等差数列前n项和为：Sn=na1+n(n-1)d/2=110.a1,a2,a4成等比数列,则a2^2=a1a4,即(a1+d)^2=a1(a1+3d),解此式得：a1=d.代入Sn式中得：na1+n(

a5=a1+4d,a13=a1+12d则有2a1+16d=34①S3=a1+a2+a3=3a1+3d=9②联立解得a1=1d=2an=2n-1Sn=n²b1=1/（1+t）b2=3/(3+t

sn=2n^2-3nS(n-1)=2(n-1)^2-3(n-1)两式相减得an=2n-2-3=2n-5所以是等差数列啊.但Sn不是了

因为an,Sn,an^2成等差数列所以2Sn=an^2+an2an=2Sn-2S(n-1)=an^2+an-a(n-1)^2-a(n-1)得:(an-a(n-1))(an+a(n-1))-(an+a(

S11=(a1+a11)*11/2=a6*11=0a6=0d=(a6-a3)/3=-8a1=24+16=40an=-8n+48Sn=(40-8n+48)n/2=-4n^2+44n=-4(n-11/2)

等差数列公式Sn=n(a1+an)/2或Sn=a1*n+n(n-1)d/2注：an=a1+(n-1)d185=a1*10+10*（10-1）d/214=a1+（10-1）d解得a1=5d=3an=5+

a(n)=a+(n-1)d=a+[n(n-1)-(n-1)(n-2)]d/2,s(n)=a(1)+a(2)+a(3)+...+a(n-1)+a(n)=na+(d/2)[0-0+2*1-0+3*2-2*

s1=as2=2a+ds3=3a+3ds2^2=s1*s3(2a+d)^2=a*(3a+3d)4a^2+4ad+d^2=3a^2+3ada^2+ad+d^2=0再解方程求出d.

/>由等差数列公式可得：a4-a2=2d=8------------d=4S10=10a1+[10x(10-1)]x4/2=190------------a1=1故等差数列通项公式为：an=a1+(n

这样应该没错,图片有点黑,不过手机应该能看得清