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.

(1)∵等比数列{an}中,a3=3,a6=24∴q3=a6a3=648=8,∴q=2;∴an=a3qn?3=2n.(2)∵a3=8,a6=64∴b3=8,b5=64,设等差数列{bn}的公差为d,∴d=b5?b35?3=28,∴bn=b3+(n-3)d=28n-76,Sn=b1n+n(n?1)2d=?48n+n(n?1)2×28=14n2-62n

设等比数列{an}的公比为q,因为a3+a6=36,①a4+a7=18 ②,②①可得a4+a7a3+a6=q=12,故a3+a6=a1q2+a1q5=14a1+132a1=36,解得a1=27,故通项公式an=27×(12)n?1=28-n,令28-n=12=2-1,解得n=9

解: 由等比中项性质得:a7²=a3·a11 a3=1,a7=5 a11=a7²/a3=5²/1=25 a11的值为25 本题考查的是等比中项性质,不需要求出首项和公比。

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(1)设公差为d,则∵S4=14,且a1,a3,a7成等比数列∴4a1+6d=14,(a1+2d)2=a1(a1+6d)∵d≠0,∴d=1,a1=2,∴an=n+1,sn=n(2+n+1)2=n(n+3)2.(2)1an?an+1=1(n+1)(n+2)=1n+1-1n+2 ∴Tn=12?13+13?14+…+1n+1?1n+2=12-1n+2=n2(n+2)∵Tn≤λan+1对任意的...

(1)设等比数列{an}的公比为q,则a1+a1q2=10a1+a1q+a1q2+a1q3=40∴a1=1q=3.∴an=a1qn-1=3n-1.∴等比数列{an}的通项公式为an=3n-1.(2)设等差数列{bn}的公差为d,则T3=b1+b2+b3=3b2=15,∴b2=5.又∵a1+b1,a2+b2,a3+b3成等比数列,∴(a2+b2...

(Ⅰ)由条件知a2-a3=2(a3-a4).(2分)即a1q-a1q2=2(a1q2-a1q3),又a1?q≠0.∴1-q=2(q-q2)=2q(1-q),又q≠1.∴q=12.(4分)∴an=64?(12)n?1=(12)n-7.(6分)(Ⅱ)bn=log2an=7-n.{bn}前n项和Sn=n(13?n)2.∴当1≤n≤7时,bn≥0,∴Tn=Sn...

(1)a4=a1*q^3 64=(-1)*q^3 q^3=-64=(-4)^3 q=-4 S4=a1*(q^4-1)/(q-1) =(-1)*[(-4)^4-1]/(-4-1) =51 (2)a3=a1q^2=3/2 (1) s3=a1+a1q+a1q^2=9/2 (2) 由 (1)得:a1=3/(2q^2) [3/(2q^2)](1+q)=9/2-3/2=3 -2q^2+q+1=0 q=1,或q=-1/2 代入得a1=3/2...

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