各项都是正数的等比数列an中,a21a 2a3,a1成等差数列
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(1)已知a3=4S3=a1+a2+a3---->a1+a2=7-4=3a2*a2=a1*a3------>4a1=a2*a2由1.2可求得a2=2或者a=-6题目已知数列{an}是各项都是正数的等比
正数项等比数列an/an-1=q,q>0根号an/根号an-1=根号q,所以{根号an}仍是等比数列.
是原数列是a1a1qa1q^2a1q^3a1q^4.根号an根号a1(根号a1)*(根号q)(根号a1)*q(根号a1)*(根号q)*q.任意相邻两项比值为是根号q因为原来q是等比数列公比,根号q不会
是{an}是各项均为正数的等比数列q大于0{根号an}是以根号a1为首项根号q为公比的等比数列
{an}为等比,各项均为正数,则:q>0a5=a3q²,a6=a3q³a3,a5,a6成等差数列则:2a5=a3+a6即:2a3q²=a3+a3q³约去a3得:
由已知an>0,得q>0,若q=1,则有Sn=na1=80,S2n=2na1=160与S2n=6560矛盾,故q≠1.∵a1(1−qn)1−q=80 &n
∵a1,1/2a3,2a2成等差数列∴2×1/2a3=a1+2a22即a3=a1+2a2∵{an}是等比数列,∴a1q²=a1+2a1q∴q²=1+2q,即q²-2q-1
解因为数列是等比数列,且公比为q则a2=a1qa3=a1q²又因为a1,1/2a3,2a2成等差数列所以有2*(1/2)a3=a1+2a2即a1q²=a1+2a1q即q²
设公比为q,则q>0a1,(1/2)a3,2a2成等差,则2(1/2)a3=a1+2a2a3=a1+2a2a1q²=a1+2a1qq²-2q-1=0(q+1)(q-2)=0q=-1
a1+a1q+a1q^2=141+q+q^2=7q=2,q=-3(舍去)an=2*2^(n-1)=2^nbn=logan=nS20=(1+20)*20/2=210
(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
设{an}的公比为q(q>0),由a3=a2+a1,得q2-q-1=0,解得q=1+52.∴a4+a5a3+a4=q=1+52.故选B.
因为已知正项等比数列{an}满足:a7=a6+2a5,则有a1q6=a1q5+2a1q4.即:q2-q-2=0,解得:q=2,q=-1,又因为时正项等比数列故q=2.∵存在两项am, an(
因为等比数列的公比q=2,则由a2a12=16,得a22q10=16,即210a22=16,解得a22=126,因为等比数列{an}的各项都是正数,所以a2=18.则a9=a2q7=18×27=16.
2a3=a2+a52a₁q²=a₁q+a₁q⁴q⁴-2q²+q=0q(q-1)(q²+q-1)=0q≠0,q≠
(1)假设存在正然数i、k、m,使得ai+ai+m=2ai+kai>0,an为等比数列,∴1+q^m=2q^k0<q<0.5而1+q^m>1>2q>2q^k∴假设不成立,an中不存在三项成等差数列.(
设公比为q,则∵各项都是正数的等比数列{an}中,3a1,12a3,2a2成等差数列,∴a3=3a1+2a2,∴q2=3+2q,∵q>0,∴q=3,∴a2012+a2014a2013+a2011=a2
1.因为(a4+a8)*(a4+a8)=a4*a4+a8*a8+2*a4*a8=a3*a5+a6*a10+2*a4*a8=41+2*4=49所以a4+a8=72.a3=a1*q的2次a10=a1*q的
^代表什么的几次方a1=1,设等比为q且q〉0,则a1+a1*q+a1*q^2=14即a1*(1+q+q^2)=14将a1代入得q^2+q-6=0解得q=-3(舍去)q=2通过验证an=2*2^n-1