已知an是等比数列,公差d不为0,若a1a3a7 2a1 a2=1
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1.1)设a1=b1=t,依题意t+3d=t*d^3,(1)t+9d=t*d^9,(2)由(1)3d/t=d^3-1代入(2)有d^9-3d^3+2=0(d^3+2)*(d^3-1)^2=0于是d^3
已知﹛an﹜是首项为-16,公差不为0的等差数列,其前n项和为Sn,且a1,a5,a4成等比数列,求﹛an﹜的公差d.设a1=-16a5=-16+4da4=-16+3d所以,a5²=a1*a
显然有:an=a1+(n-1)d,bn=b1*q^(n-1),又a3=b3,a7=b5,所以:a1+2d=a1*q^2,①a1+6d=a1*q^4,②由上面2个式子,得到:3①-②:2a1=a1*(3
a2=a1+d,a3=a1+2d.,a6=a1+5d,...,a10=a1+9d,若a1,a3,a6成等比数列,则a3^2=a1*a6,(a1+2d)^2=a1*(a1+5d),得到a1=4d.则(a
(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=
因为a5=a1+4d,a9=a1+8d,a15=a1+14d且a5a9a15成等比数列所以(a1+8d)^2=(a1+4d)(a1+14d)即(a1)^2+16a1*d+64d^2=(a1)^2+18
6m+7=3k+16(m+1)=3kk=2m+2q=bn/bn-1=an+1/an-1an+1-(an-1)=2d两个联立an-1=1+2d/q是常数所以an是常数列bn也是常数列,且bn=1
(1)∵数列{an}是公差不为零的等差数列,a1=2,且a2,a4,a8成等比数列,∴(2+3d)2=(2+d)(2+7d),解得d=2,∴an=2n.(2)∵an=2n,∴3an=32n=9n,此数
(1)a3=a1+2d、a6=a1+5d.(a1+2d)^2=a1(a1+5d)a1^2+4a1d+4d^2=a1^2+5a1d4a1d+4d^2=5a1d因为d0,所以4a1+4d=5a1a1=4d
(1).由a(m)+a(m+1)=a(k)知道3m+3(m+1)+1=3k+1,整理后有k-2m=4/3,而m,k均是N+,则k-2m也是整数,故而不存在m,k∈N+,使a(m)+a(m+1)=a(k
(1)因为a4,a5,a8成等比数列,所以a52=a4a8.设数列{an}的公差为d,则(3+3d)2=(3+2d)(3+6d)化简整理得d2+2d=0.∵d≠0,∴d=-2.于是an=a2+(n-2
解a1=1a2=1+da5=1+4da1a2a5成等比所以(1+d)^2=1*(1+4d)d^2-2d=0d=2d=0(舍)所以an=a1+(n-1)d=1+(n-1)*2=2n-1
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
设数列{an}是公差为d,且d≠0,因为a5,a10,a20三项成等比数列,所以(a1+9d)2=(a1+4d)(a1+19d),整理得5a1d=5d2,解得d=a1,则公比q=a10a5=a1+9d
(1)因为等差数列{an}的首项a1=1所以a2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d因为{bn}为等比数列所以(b3)^2=b2*b4又a2=b2,a5=b
a2,a5,a14是等比数列所以(a5)^2=a2*a14即(a+4d)^2=(a+d)*(a+13d)化简得d=2a所以公比q=a5/a2=(a+4*2a)/(a+2a)=3(2)a122=a+12
(I)设等差数列{an}的公差为d,由题意知d为非零常数∵a1=1,a1、a3、a9成等比数列∴a32=a1×a9,即(1+2d)2=1×(1+8d),解之得d=1(舍去0)因此,数列{an}的通项公
由a/an=bn,得a/a=b,{an}是公差为d的等差数列,{bn}是公比是q的等比数列,∴a/a*an/a=q,即[a1+(n-1)d][a1+(n+1)d]/[a1+nd]^2=q(常数)对n∈