高考第一輪文科數(shù)學（人教A版）課時規(guī)范練30　等比數(shù)列

課時規(guī)范練30等比數(shù)列基礎(chǔ)鞏固組1.在等比數(shù)列{an}中,a3a7=9,則a5=()A.±3 B.3C.±3 D.32.已知數(shù)列{an}為等比數(shù)列,則“a6>a5>0”是“數(shù)列{an}為遞增數(shù)列”的()A.充要條件B.充分不必要條件C.必要不充分條件D.既不充分也不必要條件3.已知等比數(shù)列{an}的前n項和為Sn,若S3=21,a4-a1=21,則a3=()A.9 B.10 C.11 D.124.在等比數(shù)列{an}中,a1+a2=10,a3+a4=20,則a7+a8=()A.80 B.100C.120 D.1405.將數(shù)列{3n+1}與{9n-1}的公共項從小到大排列得到數(shù)列{an},則a10=()A.319 B.320 C.321 D.3226.在等比數(shù)列{an}中,a1+a2=94,a4+a5=18,則其前5項的積為(A.64 B.81C.192 D.2437.(2022江蘇淮陰中學檢測)設(shè){an}是首項為1的等比數(shù)列,若a1,2a2,4a3成等差數(shù)列,則通項公式an=.

8.在等比數(shù)列{an}中,a1+a3=10,a2+a4=-5,則公比q=;滿足an>1的n的最大值為.

9.(2023四川瀘縣模擬)設(shè)Sn為數(shù)列{an}的前n項和,已知a1=1,an=2an-1+1(n≥2).(1)證明:{an+1}為等比數(shù)列;(2)求{an}的通項公式,并判斷n,an,Sn是否成等差數(shù)列?綜合提升組10.(2022廣東廣州三模)在等比數(shù)列{an}中,a1=1,且4a1,2a2,a3成等差數(shù)列,則ann(n∈N*)的最小值為(A.1625 B.49 C.111.已知數(shù)列{an}的前n項和為Sn,且滿足an+Sn=1,則S1a1+S2a12.在等比數(shù)列{an}中,a2=2,a5=14,則滿足a1a2+a2a3+…+anan+1≤212成立的n的最大值為創(chuàng)新應(yīng)用組13.(2022山西臨汾考前適應(yīng)訓練一)已知{an}為等比數(shù)列,a1=16,公比q=12.若Tn是數(shù)列{an}的前n項積,則當Tn取最大值時n為()A.4 B.5C.4或5 D.5或6

參考答案課時規(guī)范練30等比數(shù)列1.A由等比數(shù)列的性質(zhì),可得a52=a3a7=9,則a5=2.B設(shè)數(shù)列{an}的公比為q.充分性:當a6>a5>0時,q=a6a5>1,且a1=a5q4>0,則數(shù)列{an}為遞增數(shù)列;必要性:當數(shù)列{an}為遞增數(shù)列時,若a1<0,0<q<1,則0>a3.D設(shè)等比數(shù)列{an}的公比為q,顯然q≠1,則a1(1-q3)1-q=21,a1q34.A設(shè)等比數(shù)列{an}的公比為q,則q2=a3+a4a1+a2=2,∴a7+a8=(a1+a25.B由題意知,數(shù)列{an}是首項為9,公比為9的等比數(shù)列,所以an=9n,則a10=910=320.6.D設(shè)等比數(shù)列{an}的公比為q,由題意a4+a5a1+a2=q3=8,解得q=2,所以a1+a2=a1+2a1=94,所以a1=34,所以a1a2a3a4a5=a17.12n-1由a1,2a2,4a3成等差數(shù)列,得4a2=a1+4a3,∴4q=1+4q2,解得q=12,∴an=1·12n-1=12n-1.8.-123因為a1+a3=10,a2+a4=-5,所以q=a2+所以a1+a3=a1+q2a1=10,即a1=8,所以an=a1qn-1=8×-12n-1,所以當n為偶數(shù)時,an<0;當n為奇數(shù)時,an=8×-12n-1=8×12n-1=24-n>0.要使an>1,則4-n>0且n為奇數(shù),即n<4且n為奇數(shù),所以n=1或n=3,n的最大值為3.9.(1)證明由an=2an-1+1得an+1=2(an-1+1),又a1+1=2,∴數(shù)列{an+1}是以2為首項,2為公比的等比數(shù)列.(2)解由(1)得an+1=2·2n-1=2n,∴an=2n-1;∴Sn=(21+22+23+…+2n)-n=2(1-2n)∴2an=2n+1-2,n+Sn=2n+1-2,即2an=n+Sn,∴n,an,Sn成等差數(shù)列.10.D設(shè)等比數(shù)列{an}的公比為q,由于4a1,2a2,a3成等差數(shù)列,∴4a2=4a1+a3,即4a1q=4a1+a1q2,則q2-4q+4=(q-2)2=0,解得q=2,∴an=2n-1,∴ann=2n-1n,a11=1,a22=1,a33=43,a44=2,a55=165,∴an+1n+1÷ann=2nn+1×11.502因為an+Sn=1,所以當n≥2時,an-1+Sn-1=1,兩式相減,可得an-an-1+(Sn-Sn-1)=2an-an-1=0,即an=12an-1(n當n=1時,可得a1+S1=2a1=1,解得a1=12所以數(shù)列{an}表示首項為12,公比為12的等比數(shù)列,所以an=12nSn=12[1-(12)

所以Snan=1所以S1a1+S2a2+S3a3+…+S8a8=(2+22+…+28)-(1+112.3已知{an}為等比數(shù)列,設(shè)其公比為q,由a5=a2q3得,2q3=14,q3=1解得q=12,又a2=2,∴a1=4∵an+1an+2anan+1=q2=14,∴數(shù)列{anan+1}也是等比數(shù)列∴a1a2+a2a3+…+anan+1=323(1-4-n)≤212,從而有14n≥