www.1862.net > 已知{An}是各项都为正数的等比数列,Sn是其前n项和...

已知{An}是各项都为正数的等比数列,Sn是其前n项和...

由题意可设{an}的公比为q,q>0∵5S2=S4,∴5(a1+a2)=a1+a2+a3+a4,∴4(a1+a2)=a3+a4,即4(a1+a2)=(a1+a2)q2,∴q2,=4,解得q=2,∴a5=a1q4=1×24=16故选:C

设等比数列{an}的公比为q,(q>0)由题意可得2×12a3=3a1+2a2,即a1q2=3a1+2a1q,即q2-2q-3=0解之可得q=3,或q=-1(舍去)故S11?S9S7?S5=a10+a11a6+a7=a6q4+a7q4a6+a7=q4=81故答案为:81

(1)∵S4=S3+8,∴a4=S4-S3=8,又∵a3=4,∴q=a4a3=84=2,a1=a3q2=1∴等比数列{an}的通项公式an=2n-1.(2)由(1)知:数列{an}的首项为1,公比为2,an=2n-1,bn=n?2n-1,∴bn=n?2n-1,∴Tn=b1+b2+b3…+bn=1+2?2+3?22+…+n?2n-12Tn=2+2?22+3?23+…+...

(I)设{an}的公差为d,{bn}的公比为q,则依题意有q>0,∵a1=b1=1,a3+b5=21,a5+b3=13,∴1+2d+q4=211+4d+q2=13,解得d=2,q=2.∴an=1+(n-1)d=2n-1,bn=2n?1,(Ⅱ)由(I)得,an?bn=(2n-1)?2n-1,Sn=1?20+3?21+…+(2n-1)?2n-12Sn=1?2+3...

(1)解:由a2n-1,a2n,a2n+1成等差数列,可知a1,a2,a3成等差数列,a3,a4,a5成等差数列,由a2n,a2n+1,a2n+2成等比数列,可知a2,a3,a4成等比数列,则2=a1+a3a32=a2a42a4=a3+a5,又a2=1,a5=3,∴2=a1+a3a32=a42a4=a3+3,则2a32=a3...

S1=3a1-1 ===>a1=1/2 an=Sn-S(n-1) =3an-1-3a(n-1)+1 化简an/a(n-1)=3/2 所{an}是以a1=1/2为首相q=3/2为公比的等比数列 an=(1/2)*(3/2)^(n-1)

设a2=a1*q,a3=a1*q^2 a1(1+q^2)=10, a1q(1+q)=6 (1+q^2)/(q+q^2)=10/6=5/3 跳过几步,可以解得 a1=8,q=1/2

a4*a10=(a7)^2=(2a6)^2=16 所以 a6=2

解: (1) 令q=1,p=n a(n+1)=an+a1 a(n+1)-an=a1,为定值,数列{an}是以a1为首项,a1为公差的等差数列。 an=a1+a1(n-1)=na1 设{bn}公比为q,数列{bn}各项均为正,则首项b1>0,q>0 a3+b5=19 a5+b3=9 3a1+b1q⁴=19 5a1+b1q²=9 b1=1代入...

解: 等比数列各项均为正,则首项a1>0,公比q>0 (a2+a3)/(a1+a3)=6/10 (a1q+a1q²)/(a1+a1q²)=3/5 (q+q²)/(1+q²)=3/5 2q²+5q-3=0 (q+3)(2q-1)=0 q=-3(舍去)或q=½ q=½代入a1+a3=10 a1(1+q²)=10 a1=10/(1+...

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