www.1862.net > 在等比数列{An}中,A1+A7=65,A3A5=64,且An+1<An...

在等比数列{An}中,A1+A7=65,A3A5=64,且An+1<An...

(1)设数列{an}的公比为q.由等比数列性质可知:a1a7=a3a5=64,而a1+a7=65,an+1<an.∴a1=64,a7=1,(3 分)由64q6=1,得q=12,或q=-12(舍),(5 分)故an=27?n.(7 分)(2)等比数列{an}中,∵a1=64,q=12,∴S5=64×[1?(12)5]1?12=124...

(1) a1(1+q^6)=65 a1^2*q^6=64 由a(n+1)

在正项等比数列{an}中,∵a3?a5=64,∴a3?a5=a1?a7=64,∴a1+a7≥2a1?a7=264=2×8=16,当且仅当a1=a7=4时,取等号,∴a1+a7的最小值为16,故答案为:16

∵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.

在等比数列中,a1a5=a 23,∵a1=1,a5=16,∴a 23=1×16=16,即a3=±4,∵a3=a1q2>0,∴a3=4,故答案为:4.

(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-...

设等差数列{an}的公差为d,由a1+1,a3+3,a5+5构成等比数列,得:(a3+3)2=(a1+1)(a5+5),整理得:a32+6a3+4=a1a5+5a1+a5,即(a1+2d)2+6(a1+2d)+4=a1(a1+4d)+5a1+a1+4d.化简得:(d+1)2=0,即d=-1.∴q=a3+3a1+1=a1+2d+3a1+1=a1+2×(?1)+3a1...

a1+a2=1、a3+a4=9 又: (a3+a4)/(a1+a2)=q² 得: q²=9,即:q=3 而: (a4+a5)=(a3+a4)q=27

(1)a3与a5的等比中项a4=2, ∴a1a5+2a3a5+a2a8 =a4^2(1/q^2+2+q^2) =25, ∴1/q^2+2+q^2=25/4, ∴q^4-(17/4)q^2+1=0, q∈(0,1), ∴q^2=1/4,q=1/2, ∴an=a4*q^(n-4)=1/2^(n-5). (2)bn=logan=5-n, ∴Sn=n(9-n)/2. (3)S1/1+S2/2+...+Sn/n =(1/4)n(17-n)(...

a7=8

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