在公差不为零的等差数列an和等比数列 a1=b1=1,a2=v2,a8=b3
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1.因为等差数列AN的公差d不等于0,a1=2,s9=36,所以36=9*2+1/2*9*8d所以d=1/2所以a3=3,a9=6,由a3,a9,am成等比数列则a9的平方=a3*am,的am=12又
(Ⅰ)设公差为d,由条件得5a1+5×42d=30(a1+2d)2=a1(a1+8d),得a1=d=2.∴an=2n,Sn=2n+n(n-1)×22=n2+n;(Ⅱ)∵1Sn+an+2=1n2+n+2
a2=1+d,a6=1+5d由于a1、a2、a6是等比数列{bn}的前三项,所以1+5d=(1+d)^2,得d=3(公差不为零),因此{bn}的公比为q=4,故an=1+(n-1)*3=3n-2;bn
(1)a3=a1+2d,a7=a1+6d,所以a1*a7=a3*a3,即a1*(a1+6d)=(a1+2d)*(a1+2d)解得d=1(2)Sn=(1/2)n^2+(3/2)n,又a3=a1+2d=4
a1+q^2*a1=2*q*a1解得q=1不存在满足条件的答案……你检查题目是不是有问题……
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
a1*a5=a3*a4a1*(a1+4d)=(a1+2d)(a1+3d)d=2Sn最小即an
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
【第(1)题】设{an}首项为a1,公差为d(d≠0);{bn}首项为b1,公比为q(q≠0,q≠1)则,an=a1+(n-1)d,bn=b1*q^(n-1)由题意,a1=b1=1则有1+d=1*q1
(I)由题意可得,4a1+6d=10(a1+2d)2=(a1+d)(a1+6d)∵d≠0∴a1=−2d=3∴an=3n-5(II)∵bn=2an=23n-5=14•8n−1∴数列{an}是以14为首项
(1)由题意可得(a1+d)2+(a1+2d) 2=(a1+3d)2+(a1+4d)27a1+21d=7联立可得a1=-5,d=2∴an=-5+(n-1)×2=2n-7,sn=−5n+n(n
由a1+a3=4知a1+(a1+2d)=4即a1+d=2,又a2,a3,a5成等比数列得到a32=a2a5即(a1+2d)2=(a1+d)(a1+4d),a12+4da1+4d2=a12+5da1+4
/>设等差数列的公差为d(d≠0)a1.a3.a7依次成等比数列∴a3²=a1*a7∴(a1+2d)²=a1(a1+6d)∴4d²=2a1d∵d≠0∴a1=2d∴an=a
设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9
数列{an}是公差不为0的等差数列,设公差为d,S1,S2,S4成等比数列,则S22=S1•S4,∴( 2a1+d)2=a1•(4a1+6d),化简可得d=2a1∴a3a1=a1+2da1=
设a2=b2=x则a5=4x-3b3=x^2所以4x-3=x^2解得x=1(舍去,因为公差不为0)或者3所以(1)an=2n-1bn=3^(n-1)(2)S(bn)=(3^n-1)/2(3)若成立则2
设公差d,公比p.所以an=1+(n-1)d,bn=p^(n-1)所以两个方程:1+d=p,1+7d=p*p.1式带入2式,d=5,p=6
a2=a1+da3=a1+2da6=a1+5d由等比数列性质(a1+2d)^2=(a1+d)(a1+5d)a1=-1/2dq=a3/a2=3
a2,a3,a6组成等比数列的连续三项∴a3的平方=a2a6(a1+2d)²=(a1+d)(a1+5d)化简得d=-2a1q=a3/a2=(a1+2d)/(a1+d)=(-3a1)/(-a1