在公差不为零的等差数列an中,a1,a3,a7依次
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(1)∵等差数列{an}中,a2,a4,a9成等比数列,∴a42=a2•a9,即(a1+3d)2=(a1+d)(a1+8d),整理得:6a1d+9d2=9a1d+8d2,即d2=3a1d,∵d≠0,∴
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
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
2a3+2a11=4a7令a7=x-a7²+4a7=0解的a7=0(舍去),a7=4b6b8=b7²因为b7=a7结果为16再问:a7为什么不为0?再答:因为b7不能为0
an等差,则a3+a11=2a72a3-a7的平方+2a11=0→4a7-a7的平方=0bn等比,则bn不为零,即a7=b7,即a7不为零所以a7=4=b7若是求b5b8,则条件不足若是求b6b8,则
(1)设等差数列{an}的公差为d(d≠0),由a1,a3,a13成等比数列,得a32=a1•a13,即(1+2d)2=1+12d得d=2或d=0(舍去).故d=2,所以an=2n-1(2)∵bn=2
/>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
(1)设数列的公差为d,则∵a3=7,又a2,a4,a9成等比数列.∴(7+d)2=(7-d)(7+6d)∴d2=3d∵d≠0∴d=3∴an=7+(n-3)×3=3n-2即an=3n-2;(2)∵bn
ak=a1+(k-1)d=9d+(k-1)d=(k+8)da2k=a1+(2k-1)d=9d+(2k-1)d=(2k+8)d又a1a2k=ak^2,即9d(8+2k)d=[(8+k)d]^2k=4
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
∵a1=81,d=-7,∴an=81+(n-1)×(-7)=88-7n,由an=88-7n≥0,解得n≤1247,∴最接近零的是第13项,故选C.
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
依题意可知ma1+m(m−1)d2=na1+n(n−1)d2,整理得a1+n+m−12d=0∴Sm+n=(m+n)a1+(n+m−1)(n+m)2d=(n+m)(a1+n+m−12d)=0故答案为:0
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