已知等比数列an各项都是正数 ,且a1,1 2a3,2a2成等差数列 ,则等于
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由题意易知a3=a1+2a2a1*q^2=a1+2a1*q(a1不等于0)即q^2-2q-1=0,解得q=1+√2或-1+√2(√2指根号2)(a9+a10)/(a7+a8)=a9(1+q)/[a7(
设等比数列{an}的公比为q,∵各项都是正数,且a1,12a3,2a2成等差数列,∴a3=a1+2a2,即a1q2=a1+2a1q,解得q=1+2,或q=1-2(舍去).∴a6+a7a8+a9=a6+
(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
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
由a1,1/2a3,2a2成等差数列则有a3=a1+2a2,设等比数列(an)公比为q则有a2*q=a2/q+2a2因为数列(an)中各项都是正数所以两边同除以a2得q=1/q+2解得q=1+根号2或
∵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
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(
(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.(
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数列求
为等差吧{an}是等比数列所以an^2=an+1×an-1lgan^2=lg(an+1×an-1)2lgan=lgan+1+lgan-1{lgan}是等差数列Lga1+…lga10=lg(a1×.a1
(1)(Xn)^an=(Xn+1)^an+1=(Xn+2)an+2=k得Xn=k^(1/an),X(n+1)=k^(1/a(n+1)),X(n+2)=k^(1/(an+2))由等比数列{Xn}可知:(
^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