考點(diǎn)規(guī)范練27等比數(shù)列及其前n項(xiàng)和1.已知等比數(shù)列{an}的前3項(xiàng)和為168,a2-a5=42,則a6=()A.14 B.12 C.6 D.3答案:D解析:設(shè)公比為q(q≠0),則a1+a2+a3=a1(1+q+q2)=168,a2-a5=a1q-a1q4=a1q(1-q)·(1+q+q2)=42,∴q(1-q)=14,解得q=1∴a1=96,∴a6=a1q5=96×125=3.故選2.在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a2,a48是方程2x2-7x+6=0的兩個(gè)根,則a1·a2·a25·a48·a49的值為()A.212 B.93 C.±93 D.3答案:B解析:∵a2,a48是方程2x2-7x+6=0的兩個(gè)根,∴a2·a48=3.又a1·a49=a2·a48=a252=3,且a25∴a1·a2·a25·a48·a49=a255=9故選B.3.已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),且a2a4=16,a4+a5=24,則數(shù)列{an}的通項(xiàng)公式為()A.an=2n-1 B.an=2nC.an=3n D.an=3n-1答案:A解析:設(shè)等比數(shù)列{an}的公比為q.由題意可知,對(duì)任意的n∈N*,an>0,q>0.由等比中項(xiàng)的性質(zhì)可得a2·a4=a32=16,解得a3=所以a4+a5=a3(q+q2)=4(q+q2)=24,整理可得q2+q-6=0,解得q=2(負(fù)值舍去),所以an=a3qn-3=4×2n-3=2n-1.4.(2023新高考Ⅱ,8)記Sn為等比數(shù)列{an}的前n項(xiàng)和,若S4=-5,S6=21S2,則S8=()A.120 B.85 C.-85 D.-120答案:C解析:若q=1,則S6=6a1,S2=2a1,不滿(mǎn)足S6=21S2,所以q≠1.由S6=21S2得a1(1-q6)1-q=21·a1(1-q2)1-q,所以1-q6=(1-q2)(1+q2+q4)=21(1-q2),則由已知S4=a1(1-q4)1-q=-5,則5.現(xiàn)有一石窟的某處“浮雕像”共7層,上一層的數(shù)量是下一層的2倍,總共有1016個(gè)“浮雕像”,這些“浮雕像”構(gòu)成一幅優(yōu)美的圖案.若從最下層往上“浮雕像”的數(shù)量構(gòu)成一個(gè)數(shù)列{an},則log2(a3·a5)的值為()A.8 B.10 C.12 D.16答案:C解析:根據(jù)題意可知,最下層的“浮雕像”的數(shù)量為a1,且數(shù)列{an}為公比q=2的等比數(shù)列.由題意知數(shù)列{an}的前7項(xiàng)和S7=a1(1-解得a1=8,則an=8×2n-1=2n+2(1≤n≤7),于是a3=25,a5=27,從而a3·a5=25×27=212,可得log2(a3·a5)=log2212=12,故選C.6.(多選)已知數(shù)列{an}是等比數(shù)列,那么下列數(shù)列一定是等比數(shù)列的是()A.1an BC.{an+an+1} D.{an+an+1+an+2}答案:AD解析:當(dāng)an=1時(shí),log2an2=0,數(shù)列{log2an當(dāng)公比q=-1時(shí),an+an+1=0,數(shù)列{an+an+1}不一定是等比數(shù)列;由等比數(shù)列的定義知數(shù)列1an和{an+an+1+an+2}7.若a,b是函數(shù)f(x)=x2-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可通過(guò)適當(dāng)排序后成等差數(shù)列,也可通過(guò)適當(dāng)排序后成等比數(shù)列,則p+q的值等于()A.6 B.7 C.8 D.9答案:D解析:∵a,b是函數(shù)f(x)=x2-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),∴a+b=p,ab=q.∵p>0,q>0,∴a>0,b>0.又a,b,-2這三個(gè)數(shù)可通過(guò)適當(dāng)排序后成等差數(shù)列,也可通過(guò)適當(dāng)排序后成等比數(shù)列,∴2b=a-解①得a=4,b=1∴p=a+b=5,q=1×4=4.∴p+q=9.故選D.8.(多選)已知數(shù)列{an}是等比數(shù)列,則下列說(shuō)法中正確的是()A.數(shù)列{an2B.若a3=2,a7=32,則a5=±8C.若a1<a2<a3,則數(shù)列{an}是遞增數(shù)列D.若數(shù)列{an}的前n項(xiàng)和Sn=3n-1+r,則r=-1答案:AC解析:由于數(shù)列{an}是等比數(shù)列,則an2=a12q2n-2,可得an+12an2=a若a3=2,a7=32,則a5=2×32=8,故B若0<a1<a2<a3,則q>1,數(shù)列{an}是遞增數(shù)列;若a1<a2<a3<0,則0<q<1,數(shù)列{an}是遞增數(shù)列,故C正確;若數(shù)列{an}的前n和Sn=3n-1+r,則a1=S1=1+r,a2=S2-S1=(3+r)-(1+r)=2,a3=S3-S2=(9+r)-(3+r)=6,由于a1,a2,a3成等比數(shù)列,則a22=a1a3,即4=6(1解得r=-13,故D錯(cuò)誤9.已知數(shù)列{an}是等比數(shù)列,且an>0,a2a4+2a3a5+a4a6=25,則a3+a5=,a4的最大值為.

答案:55解析:因?yàn)閧an}是等比數(shù)列,a2a4+2a3a5+a4a6=25,所以a32+2a3a5+a52=25,因?yàn)閍n>0,所以a3+a5=5,故a3+a5≥2a3a5=2a4,即a410.已知等比數(shù)列{an}與等差數(shù)列{bn}滿(mǎn)足a1=b1=1,a1≠a2,a1,a2,b3成等差數(shù)列,b1,a2,b4成等比數(shù)列.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)Sn,Tn分別是數(shù)列{an},{bn}的前n項(xiàng)和,若Sn+Tn>100,求n的最小值.解:(1)設(shè)等比數(shù)列{an}的公比為q,