www.1862.net > 在等比数列{An}中,A3=1,A4=5/2,求A7. 要过程

在等比数列{An}中,A3=1,A4=5/2,求A7. 要过程

∵a2*a3=a1*a4∵a1*a4=2a1∴a4=2∵a4与2a7的等差中项为5/42*5/4=a4+2a7∴a7=1/4∵q³=a7/a4=1/8∴q=1/2∴a1=a4/q³=16∴S5=a1*(1-q^5)/(1-q)=31所以选C

设公比是q,则:a1+a3=5,a2+a4=10可写成: a1+a1q^2=5, a1q+a1q^3=10, 两式相除解得:q=2,把q=2代入a1+a1q^2=5得:a1=1, 从而a2=2, a3=4,a4=8

a1=1 a2=2 a3=3 a4=4 a5=5 这个不是等比数列

(1)设{an}的公差为d,{bn}的公比为q,a1=b1=1,a2+b2=5,a3+b3=9则a1+d+b1q=5a1+2d+b1q2=9,即d+q=4…①2d+q2=8…②,②-①×2得,q2-2q=0,∴q=2,q=0(舍),代入①得d=2.∴an=1+(n-1)?2=2n-1,bn=2n-1(2)anbn=2n?12n?1∴Sn=1+32+54+78+…+2n?1...

(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)设某等差数列为{bn},则b5=a2=a1q=64q,b3=a3=64q2,b2=a4=64q3.∵b5=b2+3(b3-b2),∴64q=64q3+3(64q2-64q3),化为2q2-3q+1=0,q≠1,解得q=12.∴an=a1qn?1=64×(12)n?1=(12)5+n.(2)∵bn=log2an=log22?5?n=-n-5.∴|bn|=n+5∴数列{|bn|...

(1)公比为2所以a2=2a1 a3=4a1 a4=8a1原式=(2a1+2a1)/(8a1+8a1)=1/4 (2)设这个常数是C则(50+c)^2=(20+c)(100+c)求得c=25所以公比是 (50+25)/(20+25)=75/45=5/3

解:设等比数列的公比为q a1+a3=a1*(1+q^2)=10 (1) a4+a6=a1*(q^3+q^5)=5/4,即a1*(1+q^2)*q^3=5/4 (2) (2)/(1)得:q^3=1/8,即q=1/2 所以a1=8, 所以s4=a1*(1-q^4)/(1-q)=15, s5=a1*(1-q^5)/(1-q)=15.5

an = a1.q^(n-1) a3=-3 a1.q^2 =-3 (1) a6=-1/9 a1.q^5 =-1/9 (2) (2)^5/(1)^2 (a1)^(5-2) . q^(25-4) = (-1/9)^5/(-3)^2 (a1)^3 . q^21 = -(1/3)^12 a1.q^7 = -(1/3)^(12/3) a8 = a1.q^7 = -(1/3)^4 = -1/81

∵2a2,S3,a4+2成等差数列,a1=1∴2S3=2a2+a4+2∴q≠1∴2×1?q31?q=2q+q3+2∴q3-2q2=0∵q≠0∴q=2∴数列{an2}是以1为首项,以4为公比的等比数列前5项和为1?451?4=341故选A

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