已知数列an为等比数列a5×a2=32 9 a1 a6=11求
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Sn=a1*(1-q^n)/(1-q),Tn=a1*(1-q^2)/(1-q)q不等于1时,lim(Sn)/Tn=lim(1-q^n)/(1-q^2n)q1,lim(Sn)/Tn=lim1/q^n=0
a5^2=a10.得出(a1*q^4)^2=a1*q^9得出a1=qAn为递增数列,说明q>12[An+A(n+2)]=5A(n+1)A(n+2)=an·q^2;A(n+1)=an·q代入上式得:2A
(1)已知{an}为递增的等比数列可知等比不可能是负数,有以下2种情况若q
无解厄题目抄错了哇a5+a6+a7+a8=a1*q^4+a2*q^4+a3*q^4+a4*q^4=q^4(a1+a2+a3+a4)=-5q^4*10=-5不存在
已知公差为d(d不等于0),a1=1,那么:a2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d又a2a5a14依次成等比数列,所以:(a5)²=a2*a14
(1)设An=q^(n-1),Bn=1+(n-1)d.由所给等式得q^4+2d=20,q^2+4d=12,设m=q^2,n=2d,则m,q>0,简化式子,易解得q=b=2.得An,Bn.(2)Bn/2
10+(-5)+(-5/-2)+(-5/4)=25/4
等比数列?题目错了吧!是不是等差数列哦?这里an=a1*qn-1a1+a2+a3+a4=10a5+a6+a7+a8=a1*q4+a2*q4+a3*q4+a4*q4=q4(a1+a2+a3+a4)=-5
1.只有常数数列才能满足既成等比也成等差a10为12、等比a2+a4+.+a20=a1q+a3q+.+a19q=q(a1+a3+.+a19)=6故a1+a3+.+a19=6/3=2s20=a2+a4+
设数列的公比为q,首项为a1,则∵a52=a10,2(an+an+2)=5an+1,∴(a1q4)2=a1q9,2(1+q2)=5q,∵等比数列{an}为递增数列,∴q=2,a1=2∴an=2n故答案
设{an}的公比为q,则q^3=a5/a2=64,可求出q=4.由a2=a1*q=2,求出a1=1/2.则an=a1*q^n-1=1/2*4^n-1=2^(2n-3).则bn=log2an=2n-3.
设等比数列的公比为q由a5²=a10>0得(a1q^4)^2=a1q^9a1=q由2[an+a(n+2)]=5a(n+1)得2[an+q^2an]=5qan所以2q^2-5q+2=0解得q=
a1=a2-2,a5=a2+6∴a22=a1a5=(a2-2)(a2+6),解得a2=3故选D
an=q^(n-1),bn=1+(n-1)d可得q^4+1+2d=21,q^2+1+4d=13解得q^2=4,d=2又q>0故q=2所以an=2^(n-1),bn=2n-1
因为a1a5=20,a2+a4=-12{a}等比,所以a2a4=20,a2,a4是方程:x^2+12x+20=0的根x=-2或x=-101.a2=-2,a4=-10q^2=a4/a2=5a8=a4*q
a2=5,a5=0.25a5=a2*q^3q^3=a5/a2=0.25/5=0.05q=0.05^(1/3)a₁a₂+a₂a₃+...+an×a(n+1
(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
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)设等比数列{an}的公比为q,则a2=a1q,a5=a1q4.依题意,得方程组a1q=6a1q4=162解此方程组,得a1=2,q=3.故数列{an}的通项公式为an=2•3n-1.(2)Sn=
∵等差数列{an}中,a1,a2,a5成等比数列,∴a22=a1•a5,即(a1+d)2=a1•(a1+4d),又d=2,∴(a1+2)2=a1•(a1+8),整理得:a12+4a1+4=a12+8a