已知等差数列an的首项和等比数列bn首项相等公差公比都是d
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a(n)>0.s(n)=[a(n)+1]/4,a(1)=s(1)=[a(1)+1]/4,a(1)=1/3.s(n+1)=[a(n+1)+1]/4,a(n+1)=s(n+1)-s(n)=[a(n+1)+
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=
a3=a1+2d=a1+4a4=a1+3d=a1+6因为a1,a3,a4成等比数列,则a4/a3=a3/a1(a1+4)^2=a1(a1+6)解之,a1=-8则a2=a1+d=-8+2=-6
(a1+4d)^2=a1(a1+6d),把a1=8,带入得d=-1或d=0s10=10*8-45=35,或s10=80再答:客气
因为a3+a5+a7=9所以3*a5=9所以a5=3所以a5-a1=4d=3-1=2(d是公差)所以d=0.5所以an=a1+(n-1)d=1+(n-1)*0.5即an=0.5n+0.5字数不够bn过
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:
S5^2=S3*S4(S3+S4)/2=1=>S3+S4=2S3=a1+a1+d+a1+2d=3a1+3dS4=4a1+7dS5=5a1+12dS3+S4=7a1+10d=2=>d=1/5-7/10a
怎么会有相同的题目,刚刚答完那边那个75首先a1=5,b2=5,从这个开始{an}公差为3,{bn}公差为4,公倍数为12可以发现,对于{an}来讲每12/3=4个会有一个出现在{bn}中对于{bn}
设等差数列{an}的公差为d(d≠0),则6a1+15d=60a1a21=a62,即6a1+15d=60a1(a1+20d) =(a1+5d) 2,解得:d=2a1=5,∴an=5
设等差数列公差为d;a(2)^2=a(1)a(4)S(3)=48=a(1)+a(2)+a(3)=3a(2)a(2)=48/3=16;a(1)a(4)=a(2)^2=256=[a(2)-d][a(2)+
Sn与2的等比中项为√(2Sn),an与2的等差中项为(an+2)/2由题目可知,8Sn=(an+2)^2,所以8S_(n-1)=[a_(n-1)+2]^2.两者相减,得8an=an^2+4an-[a
简单的要死,你成绩在学校排中等吗?log2(Sn+1)=n,所以Sn+1=2^n,Sn=2^n-1,an=Sn-S(n-1)=(2^n-1)-(2^(n-1)-1)=2^(n-1)a(n+1)/an=
∵{an}为等差数列,其公差d=-2,且a7是a3与a9的等比中项,∴(a1-12)2=(a1-4)(a1-16),解得a1=20,∴S10=10a1+10×92d=110故答案为:110
设等差数列公差为d;a(2)^2=a(1)a(4)S(3)=48=a(1)+a(2)+a(3)=3a(2)a(2)=48/3=16;a(1)a(4)=a(2)^2=256=[a(2)-d][a(2)+
由题意得1S3=a1+a2+a3=7……1;6a2=a1+1+a3+6……22式+1式得a2=2……3将3式代入12得q=2或1/2a1=4或1an=4*(1/2)^(n-1)或an=2^(n-1)2
a2^2=a1*a4,由等差得an=a1+(n-1)d,a1=a;带入得,an=na;则,1/a2+1/a2的平方+1/a2的3次+…+1/a2的n次与1/a1的大小等价于1/a2+1/a2的平方+1
1求AN的通项公式2此数列是否存在三项ar,as,at(r小于s小于t)成等差an+2为等比数列.an+2=(a1+2)2^(n-1)=2^(n+1)an=2^(n+1)
首项a1=2,公差d=2ak=a1+(k-1)d=2kS(k+2)=(k+2)(a1+a(k+2))/2=(k+2)(a1+a1+(k+2-1)d)/2=(k+2)(a1+k+1)=(k+2)(k+3
由a3与a5的等比中项为2根号6,可知a3a5=24①又等差数列{an}中,a2+a6=10所以a3+a5=a2+a6=10②由①②解得:a3=4,a5=6(因为an>0)所以公差d=(a5-a3)/