www.1862.net > 已知等比数列【An】的公比为正数,前n项和为Sn,A1...

已知等比数列【An】的公比为正数,前n项和为Sn,A1...

当q=1时,Sn+1=(n+1)a1,Sn=na1,所以limn→∞Sn+1Sn=limn→∞n+1n=1成立,当q≠1时,Sn=a1(1?qn)1?q,所以limn→∞Sn+1Sn=limn→∞1?qn+11?qn,可以看出当0<q<1时,limn→∞1?qn+11?qn=1成立,故q的取值范围是(0,1].故选B.

解:设{a n }的公差为d,{b n }的公比为q,由a 3 +b 3 =17得1+2d+3q 2 =17, 由T 3 -S 3 =12得q 2 +q-d=4, 由①、②及q>0解得q=2,d=2, 故所求的通项公式为a n =2n-1,b n =3×2 n-1 。

因为等比数列{an}的各项均为正数,所以q>0,由题意得S2=3a2+2 ①,S4=3a4+2 ②,②-①得,a3+a4=3a4-3a2,则a2q+a2q2=3a2q2?3a2,即2q2-q-3=0,解得q=32或q=-1(舍去),故选:C.

∵等比数列{a n }的各项均为正数,公比q≠1,∴ P= a 3 + a 9 2 > a 3 ? a 9 = Q= a 5 ? a 7 ,故选A.

∵等比数列{a n }的首项为正数,公比大于1,则可设数列{a n }的公比为q,则q>1则a n =a 1 q n-1 ,a n+1 =a 1 q n ,令b n = log 1 3 a n ,则b n+1 -b n = log 1 3 (a 1 q n ) - log 1 3 (a 1 q n-1 ) = log 1 3 q <0故数列 { log 1 3 a n } ...

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