課時(shí)規(guī)范練27　等比數(shù)列

1、 課時(shí)規(guī)范練27等比數(shù)列基礎(chǔ)鞏固組1.(2021山東泰安高三月考)已知各項(xiàng)均為正數(shù)的等比數(shù)列an的前n項(xiàng)和為Sn,若S4-S3=27,S3-S1=12,則S5=()A.81B.812C.121D.24322.(2021遼寧沈陽(yáng)高三期中)在等比數(shù)列an中,a1+a2+a3=3,a4+a5+a6=6,則數(shù)列an的前12項(xiàng)和為()A.90B.60C.45D.323.(2021陜西高新一中高三二模)已知數(shù)列an是等比數(shù)列,Sn是其前n項(xiàng)積,若S7S2=32,則S9=()A.1 024B.512C.256D.1284.(2021云南曲靖高三月考)設(shè)等比數(shù)列an的前n項(xiàng)和為Sn,若S3,S9,S6成等差數(shù)

2、列,且a8=6,則a11=()A.-1B.-3C.-5D.-75.(2021湖南長(zhǎng)沙高三期末)在公比不為1的等比數(shù)列an中,存在s,tN*,滿足asat=a52,則4s+14t的最小值為()A.34B.712C.58D.136.在各項(xiàng)均為正數(shù)的等比數(shù)列an中,Sn是數(shù)列an的前n項(xiàng)和,若a1=12,S3=78,則下列說法錯(cuò)誤的是()A.an12B.an12C.Sn|an|,a2=-2,S3=3,則|a1|+|a2|+|an|=.綜合提升組10.(2021江蘇泰州高三月考)已知為非零常數(shù),數(shù)列an與2an+均為等比數(shù)列,且a2 021=3,則a1=()A.1B.3C.2 021D.4 04211

3、.已知數(shù)列an的前n項(xiàng)和為Sn,若a1+a2=2,an+1=Sn+1,則()A.數(shù)列an是公比為2的等比數(shù)列B.S6=49C.anSn既無最大值也無最小值D.1a1+1a2+1an1,則()A.a1a3B.a1a3D.a20).由題意S4-S3=27,S3-S1=12,可得a4=27,a2+a3=12,所以a1q3=27,a1q+a1q2=12,所以a1=1,q=3,所以S5=1(1-35)1-3=121.故選C.2.C解析: 設(shè)數(shù)列an的公比為q,則a4+a5+a6a1+a2+a3=q3=63=2,所以a7+a8+a9=(a4+a5+a6)q3=62=12,同理a10+a11+a12=24,

4、所以S12=a1+a2+a12=3+6+12+24=45.故選C.3.B解析: 因?yàn)镾7S2=a3a4a5a6a7=(a5)5=32,所以a5=2,所以S9=a1a2a3a4a5a6a7a8a9=(a5)9=512.故選B.4.B解析: 設(shè)等比數(shù)列an的公比為q,由S3,S9,S6成等差數(shù)列,可得2S9=S3+S6,即2(S9-S6)=S3-S6,即2(a7+a8+a9)=-(a4+a5+a6),即q3=-12.又a8=6,所以a11=a8q3=6-12=-3.故選B.5.C解析: 設(shè)等比數(shù)列an的公比為q(q1).因?yàn)閍sat=a52,所以a1qs-1a1qt-1=(a1q4)2,即qs+t

5、-2=q8,可得s+t=10,且s,tN*,則4s+14t=1104s+14t(s+t)=1104+4ts+s4t+14=110174+4ts+s4t.因?yàn)閟,tN*,所以4ts0,s4t0,則4ts+s4t24tss4t=2,當(dāng)且僅當(dāng)4ts=s4t,即s=8,t=2時(shí),等號(hào)成立,所以4s+14t的最小值為58.故選C.6.B解析: 設(shè)等比數(shù)列an的公比為q(q0).由a1=12,S3=78,得12+12q+12q2=78,即4q2+4q-3=0,解得q=12或q=-32(舍去),所以an=1212n-1=12n.又nN*,所以an12,故選項(xiàng)A正確,選項(xiàng)B錯(cuò)誤;Sn=121-(12)n1-1

6、2=1-12n1,故選項(xiàng)C正確;2lg an=2lg12n=lg122n=lg12n-2+lg12n+2=lg an-2+lg an+2,故選項(xiàng)D正確.故選B.7.D解析: 由S1,S2+2,S3成等差數(shù)列,可得a2=4,a1=2,an=2n,故選項(xiàng)A正確;a3=8,a4-4=12,a2,a3,a4-4成等差數(shù)列,故選項(xiàng)B正確;Sn=2n+1-2,所以Sn+2=2n+1,所以Sn+2是等比數(shù)列,故選項(xiàng)C正確;若am,an,ar即2m,2n,2r成等差數(shù)列,不妨設(shè)mn|an|,所以|q|1,所以數(shù)列an是以1為首項(xiàng),以-2為公比的等比數(shù)列,則|an|也為等比數(shù)列,公比為2,首項(xiàng)|a1|=1,故|

7、a1|+|a2|+|an|=1-2n1-2=2n-1.10.B解析: an與2an+均為等比數(shù)列,所以(2an+)2=(2an-1+)(2an+1+)且an2=an-1an+1,得2an=an-1+an+1,故數(shù)列an也為等差數(shù)列,所以數(shù)列an為非零常數(shù)列,則a1=a2 021=3.故選B.11.D解析: an+1=Sn+1,令n=1,得a2=S1+1=a1+1,結(jié)合a1+a2=2,知a1=12,a2=32.又an+1=Sn+1,所以an=Sn-1+1(n2),所以an+1=2an,但a1=12,a2=32,a2a1=32.當(dāng)n2時(shí),Sn=32(1-2n-1)1-2+12=32n-2-1,S6

8、=316-1=47,故選項(xiàng)A,B錯(cuò)誤;因?yàn)閍1S1=1,當(dāng)n2時(shí),anSn=32n-332n-2-1=36-12n-312,34,所以anSn無最小值,有最大值,故選項(xiàng)C錯(cuò)誤;1a1+1a2+1an=2+23(1-12n-1)1-120,q1),則Sn=a1(1-qn)1-q,S2n=a1(1-q2n)1-q,bn=S2nSn=1-q2n1-qn=(1+qn)(1-qn)1-qn=1+qn0,b6b2-78=1+q61+q2-78=(1+q2)(1-q2+q4)1+q2-78=q4-q2+18,令q2=t(t0且t1),則b6b2-78=t2-t+18,b6b2與78大小關(guān)系不確定,即7b2與8b6大小關(guān)系不確定,故選項(xiàng)A,選項(xiàng)B錯(cuò)誤;b9b3-34=1+q91+q3-34=(1+q3)(1-q3+q6)1+q3-34=q6-q3+14=q3-1220,即b9b334,3b34b9.又b4-b8-14=1+q4-1-q8-14=-q8+q4-14=-q4-1220,即b4b8+14,故選項(xiàng)D正確.故選D.16