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...

(1)设等比数列{an}的公比为q,∵a3=4,a7=64,∴a1q2=4a1q6=64,解得a1=1,q=2.∴an=1×2n?1=2n?1.(2)Sn=1×(2n?1)2?1=2n-1.

∵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)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...

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)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...

(1)在等比数列中,由a1+a7=65,a3?a5=64,得a1+a7=65,a3?a5=a1a7=64,解得a1=1,a7=64或a1=64,a7=1,由an+1<an得数列为递减数列,∴a1=64,a7=1,解得64q6=1,即q6=164=(12)6,解得q=12或q=?12(舍去).∴求数列{an}的通项公式问an=64?(...

等比数列{an}中,由a3a11=4a7,可知a72=4a7,∴a7=4,∵数列{bn}是等差数列,∴b5+b9=2b7 =2a7 =8,故选C.

(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-...

设等比数列{an}的公比为q,则q2=a7a5=64=32,∴a9=a7q2=6×32=9故选C

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