老高考舊教材適用2025版高考數(shù)學(xué)二輪復(fù)習(xí)專題二數(shù)列考點(diǎn)突破練4等差數(shù)列等比數(shù)列文

考點(diǎn)突破練4等差數(shù)列、等比數(shù)列一、選擇題1.(2024·北京一模)已知數(shù)列{an}的前n項(xiàng)和Sn=n2,則{an}是()A.公差為2的等差數(shù)列B.公差為3的等差數(shù)列C.公比為2的等比數(shù)列D.公比為3的等比數(shù)列2.(2024·江西上饒一中二模)在等比數(shù)列{an}中,若a5=9,則log3a4+log3a6=()A.2 B.3 C.4 D.93.等差數(shù)列{an}的前n項(xiàng)和為Sn,若?n∈N*,Sn≤S7,則數(shù)列{an}的通項(xiàng)公式可能是()A.an=3n-15 B.an=17-3nC.an=n-7 D.an=15-2n4.(2024·河北石家莊二模)等差數(shù)列{an}的前n項(xiàng)和記為Sn,若a2+a2021=6,則S2022= ()A.3033 B.4044 C.6066 D.80885.(2024·江西九江二模)若數(shù)列{an}為等比數(shù)列,且a1,a5是方程x2+4x+1=0的兩根,則a3=()A.-2 B.1 C.-1 D.±16.(2024·重慶二模)已知等差數(shù)列{an}的公差為2,前n項(xiàng)和為Sn,若am=5,則Sm的最大值為()A.3 B.6 C.9 D.127.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a5+2a10+a13=18,則S18=()A.74 B.81 C.162 D.1488.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S4=3,Sn-4=12,Sn=17,則n的值為()A.17 B.15 C.13 D.119.(2024·北京順義二模)設(shè)等比數(shù)列{an}的公比為q,前n項(xiàng)和為Sn.若Sn=2,n=1,qn-A.8 B.9 C.18 D.5410.若等差數(shù)列{an}和{bn}的前n項(xiàng)的和分別是Sn和Tn,且SnTn=n2A.1221 B.11C.613 D.11.我國古代數(shù)學(xué)名著《九章算術(shù)》中,有已知長方形面積求一邊的算法,其前兩步為:第一步:構(gòu)造數(shù)列1,12,13其次步:將數(shù)列①的各項(xiàng)乘以n2,得到一個(gè)新數(shù)列a1,a2,a3,…,an則a1a2+a2a3+a3a4+…+an-1an=()A.n24 BC.n(n-12.(2024·四川涼山二模)已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}與各項(xiàng)均為正數(shù)的等差數(shù)列{bn},若a1a5=b5b7,則a3與b6的關(guān)系是 ()A.a3=b6 B.a3≥b6C.a3≤b6 D.以上都不正確二、填空題13.(2024·江西二模)已知各項(xiàng)均為正數(shù)的等差數(shù)列{an}的前n項(xiàng)和為Sn,且a5=9,若3a2,a14,S9成等比數(shù)列,則數(shù)列{an}的通項(xiàng)公式為an=.

14.(2024·廣西柳州三模)已知數(shù)列{an}的前n項(xiàng)和為Sn,an+Sn=2,則a5=.

15.已知等差數(shù)列{an}的公差為2,若a1,a3,a4成等比數(shù)列,則a10=.

16.(2024·浙江寧波二模)2024年北京冬奧會(huì)開幕式以中國傳統(tǒng)二十四節(jié)氣作為倒計(jì)時(shí)進(jìn)入,草木生長的勃勃朝氣拉開春意盎然的開幕式序幕.在中國古代,人們用圭表測量日影長度來確定節(jié)氣,一年之中日影最長與最短的日子分別被定為冬至與夏至,其日影長分別為13.5尺與1.5尺.從冬至到夏至,依次有冬至、小寒、大寒、立春、雨水、驚蟄、春分、清明、谷雨、立夏、小滿、芒種、夏至這十三個(gè)節(jié)氣,其日影長依次成等差數(shù)列,則北京冬奧會(huì)開幕日(立春)的日影長為尺.

考點(diǎn)突破練4等差數(shù)列、等比數(shù)列1.A解析:當(dāng)n=1時(shí),a1=S1=1.當(dāng)n≥2時(shí),an=Sn-Sn-1=n2-(n-1)2=2n-1.a1也滿意上式,所以an=2n-1.易知an≠0.因?yàn)閍n+1-an=(2n+1)-(2n-1)=2,an+1所以該數(shù)列是以2為公差的等差數(shù)列.2.C解析:因?yàn)閧an}為等比數(shù)列,所以a52=a4a所以log3a4+log3a6=log3(a4a6)=log3a52=2log39=3.D解析:由題意可知,等差數(shù)列{an}為遞減數(shù)列,且a7≥0,a8≤0.A,C選項(xiàng)為遞增數(shù)列,故解除;對于B,a7=17-3×7=-4<0,故解除;對于D,a7=15-2×7=1>0,a8=15-2×84.C解析:因?yàn)閧an}為等差數(shù)列,所以a2+a2021=a1+a2022=6,所以S2022=2022(a1+a5.C解析:由a1,a5是方程x2+4x+1=0的兩根,得a1+a5=-4<0,a1a5=1>0,可知a1<0,a5<0.又{an}為等比數(shù)列,所以a3<0,且a32=a1a5=1,所以a3=-6.C解析:因?yàn)閍m=5,d=2,所以a1+2(m-1)=5,所以a1=7-2m(m∈N*),則Sm=m(a1+am)2又m∈N*,所以當(dāng)m=3時(shí)Sm取最大值9.7.B解析:因?yàn)閧an}是等差數(shù)列,所以a5+2a10+a13=2a9+2a10=18,即a9+a10=9,所以S18=18(a1+a18)2=9(8.A解析:∵Sn-Sn-4=an-3+an-2+an-1+an=5,S4=a1+a2+a3+a4=3,{an}為等差數(shù)列,∴(an-3+a4)+(an-2+a3)+(an-1+a2)+(an+a1)=4(a1+an)=8,∴a1+an=2,∴Sn=n(a1+a9.C解析:因?yàn)镾n=2,n=1,qn-1,n>1,故可得a1=2,a2=S2-S1=q2又?jǐn)?shù)列{an}是等比數(shù)列,則a2=a1q,即q2-3=2q,解得q=3或q=-1.若q=3,則a1=2,a2=6,a3=18;若q=-1,則a1=2,a2=-2,a3=a2q=2≠S3-S2,不滿意題意,舍去.故a3=18.10.B解析:因?yàn)榈炔顢?shù)列{an}和{bn}的前n項(xiàng)的和分別是Sn和Tn,且SnTn11.C解析:由題意知所得新數(shù)列為1×n2,1所以a1a2+a2a3+a3a4+…+an-1an=n2411×2+12×3+13×4+…+1(n-1)×n=n241-12+12-112.C解析:設(shè)等差數(shù)列{bn}的公差為d,則b5b7=(b6-d)(b6+d)=b62-d∵{an}為等比數(shù)列,∴a1a5=a32,∴a32=又{an},{bn}的各項(xiàng)均為正數(shù),∴a3≤b6.13.2n-1解析:∵{an}為等差數(shù)列,∴S9=9(a1+a設(shè){an}的公差為d,由{an}的各項(xiàng)均為正數(shù),得d>0.由題可知a142=3a2S∴(a5+9d)2=3(a5-3d)·9a5,解得d=2或d=-13(舍去),∴an=a5+(n-5)d=9+2×(n-5)=2n-1.14.116解析:當(dāng)n=1時(shí),a1=S1=2-a1,即a1=1當(dāng)n≥2時(shí),an=Sn-Sn-1=an-1-an,故2an=an-1,所以數(shù)列{an}是首項(xiàng)為1,公比為12的等比數(shù)列,即an=12n-1,則a5=11615.10解析:由題可得a3=a1+4,a4=a1+