www.1862.net > 已知等比数列{An}中,A3=4,A7=64.(1)求数列{An...

已知等比数列{An}中,A3=4,A7=64.(1)求数列{An...

∵等比数列{an}中,a3=4,a7=64,∴a1q2=4,a1q6=64,∴q=±2,a1=1,∴S7=a1(1?q7)1?q=1?271?2或1?(?2)71+2=127或43.故答案为:127或43.

a1×a9=64,所以a3×a7=64,又a3+a7=20,解得a3=4,a7=16;或a3=16,a7=4.当a3=4,a7=16时,a11=64.当a3=16,a7=4时,a11=1

(1)设等比数列的公比为q,有a1q=4a1q2+a1q3=24,解得a1=2,q=2,所以an=2n;(5分)(2)由(1)知bn=log22n=n,有an+bn=2n+n,从而Tn=(2+22+…+2n)+(1+2+…+n)=2n+1+n(n+1)2-2.(10分)

(1)q^3=a7/a4=8/2=4 q=4^(1/3)=2^(2/3) a1=a4/q^3=2/4=1/2 an=a1*q^(n-1)=1/2*(2^(2/3))^(n-1)=2^(-1+2n/3-2/3)=2^(2n/3-5/3) (2) a2+a5=a2(1+q^3)=18 a3+a6=a3(1+q^3)=9 下式/上式得:q=a3/a2=1/2 a2+a5=a1*1/2+a1*(1/2)^4=18 a1=32 an=a1*q...

解(1)设数列前6项的公差为d,d为整数,则a5=-1+2d,a6=-1+3d,d为整数,又a5,a6,a7成等比数列,所以(3d-1)2=4(2d-1),解得d=1,-------4分当n≤4时,an=n-4,由此a5=1,a6=2,数列第5项起构成以2为公比的等比数列.当n≥5时,an=2n-5,故...

(Ⅰ)设等比数列{an}的公比为q(q≠0),由a7=a1q6=1,得a1=q-6,从而a4=a1q3=q-3,a5=a1q4=q-2,a6=a1q5=q-1.…(3分)因为a4,a5+1,a6成等差数列,所以a4+a6=2(a5+1),即q-3+q-1=2(q-2+1),q-1(q-2+1)=2(q-2+1).所以q=12.故an=a1qn-...

答: a4=a3q 所以q=a4/a3=5/2 a7=a4q³=(5/2)^4=625/16

设bn=1/an+1,即bn为等差数列,b3=1/a3+1=4/3,b7=1/a7+1=2,则公差d=(b7-b3)/4=(2-4/3)/4=1/6 ∴首项b1=b3-2d=4/3-1/3=1,则bn=b1+(n-1)d=1+(n-1)/6 ∴1/an+1=bn=1+(n-1)/6 an=6/(n-1) 数列的题记住等差数列和等比数列的通项公式就可以了,根据题意列...

设数列前6项的公差为d,d为整数,由a3=-1,得:a5=a3+2d=-1+2d,a6=a3+3d=-1+3d,又a5,a6,a7成等比数列,且a7=4,所以(3d-1)2=4(2d-1),解得d=59或d=1,因为d为整数,所以d=1.所以,当n≤6时,an=a3+(n-3)×1=-1+(n-3)=n-4,由此a5=1...

(1) a1(1+q^6)=65 a1^2*q^6=64 由a(n+1)

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