2025版高考數(shù)學(xué)一輪總復(fù)習(xí)6.2等差數(shù)列習(xí)題

.2等差數(shù)列基礎(chǔ)篇固本夯基考點(diǎn)一等差數(shù)列及其前n項(xiàng)和1.(2024屆河南名校聯(lián)盟11月月考,3)在遞增的等差數(shù)列{an}中,a3+a6=-6,a4a5=8,則公差d=()A.4B.2C.-2D.2或-2答案B2.(2024課標(biāo)Ⅰ,4,5分)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若3S3=S2+S4,a1=2,則a5=()A.-12B.-10C.10D.12答案B3.(2024陜西寶雞一模,3)在1和2兩數(shù)之間插入n(n∈N*)個(gè)數(shù),使它們與1,2組成一個(gè)等差數(shù)列,則當(dāng)n=10時(shí),該數(shù)列的全部項(xiàng)的和為()A.15B.16C.17D.18答案D4.(2024屆贛州贛縣三中期中,7)已知數(shù)列{an}中,a1=1且1an+1=1an+13(n∈N*),A.14B.12C.1答案C5.(2017課標(biāo)Ⅲ,9,5分)等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則{an}前6項(xiàng)的和為()A.-24B.-3C.3D.8答案A6.(2024課標(biāo)Ⅰ,9,5分)記Sn為等差數(shù)列{an}的前n項(xiàng)和.已知S4=0,a5=5,則()A.an=2n-5B.an=3n-10C.Sn=2n2-8nD.Sn=12n2答案A7.(2024四川遂寧重點(diǎn)中學(xué)月考,5)在等差數(shù)列{an}中,a1=1,公差d≠0,其前n項(xiàng)和為Sn,若a1,a3,a13成等比數(shù)列,則lgS2S1+lgS3S2+lgS4A.1B.2C.10D.100答案B8.(2024安徽黃山重點(diǎn)中學(xué)月考,13)已知數(shù)列{an}中,a3=7,a7=3,且1an-1是等差數(shù)列,則a10答案79.(2024江蘇,8,5分)已知數(shù)列{an}(n∈N*)是等差數(shù)列,Sn是其前n項(xiàng)和.若a2a5+a8=0,S9=27,則S8的值是.

答案1610.(2024吉林樺甸四中、磐石一中等4月聯(lián)考,14)已知數(shù)列{an}、{bn}都是等差數(shù)列,其前n項(xiàng)和分別為Sn和Tn,若對(duì)隨意的n∈N*都有SnTn=7n+14答案811.(2024屆山東青島期中,16)已知Sn是數(shù)列{an}的前n項(xiàng)和,Sn=14n-n2,則an=;若Tn=|a1|+|a2|+|a3|+…+|an|,則T20=.

答案15-2n21812.(2024新高考Ⅱ,17)記Sn為公差不為零的等差數(shù)列{an}的前n項(xiàng)和,若a3=S5,a2a4=S4.(1)求{an}的通項(xiàng)公式;(2)求使得Sn>an成立的n的最小值.解析(1)設(shè)等差數(shù)列{an}的公差為d.a3=S5?a1+2d=5a1+10d?a1=-2d①,a2a4=S4?(a1+d)(a1+3d)=4a1+6d②,將①代入②得-d2=-2d?d=0(舍)或d=2,∴a1=-2d=-4,∴an=-4+(n-1)×2=2n-6.(2)由(1)知an=2n-6,Sn=na1+n(n-1)Sn>an?n2-5n>2n-6?n2-7n+6>0?(n-1)(n-6)>0,解得n<1(舍)或n>6,∴n的最小值為7.13.(2024課標(biāo)Ⅱ,17,12分)記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求{an}的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解析(1)設(shè){an}的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.∴{an}的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.∴當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.考點(diǎn)二等差數(shù)列的性質(zhì)1.(2024屆黑龍江八校期中,3)在等差數(shù)列{an}中,已知a3+a5+a7=18,則該數(shù)列前9項(xiàng)的和為()A.54B.63C.66D.72答案A2.(2017課標(biāo)Ⅰ,4,5分)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a4+a5=24,S6=48,則{an}的公差為()A.1B.2C.4D.8答案C3.(2024陜西寶雞二模,5)已知{an}是等差數(shù)列,滿意3(a1+a5)+2(a3+a6+a9)=18,則該數(shù)列前8項(xiàng)和為()A.36B.24C.16D.12答案D4.(2024江西宜春重點(diǎn)中學(xué)月考,4)在等差數(shù)列{an}中,a3,a9滿意不等式x2+24x+12<0的解集為{x|a3<x<a9},則數(shù)列{an}的前11項(xiàng)和等于()A.66B.132C.-66D.-132答案D5.(2024屆太原期中,7)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且S6=11,S9=17,則S15=()A.15B.23C.28D.30答案D6.(2024安徽滁州重點(diǎn)中學(xué)質(zhì)監(jiān),10)等差數(shù)列{an}和{bn}的前n項(xiàng)和分別為Sn和Tn,且SnTn=2n3nA.23B.79C.2031答案D7.(2024寧夏吳忠一模,7)數(shù)列{an}是等差數(shù)列,Sn為其前n項(xiàng)和,且a1<0,a2024+a2024<0,a2024·a2024<0,則使Sn<0成立的最大正整數(shù)n是()A.2024B.2024C.4040D.4041答案C8.(2024北京,8,4分)在等差數(shù)列{an}中,a1=-9,a5=-1.記Tn=a1a2…an(n=1,2,…),則數(shù)列{Tn}()A.有最大項(xiàng),有最小項(xiàng)B.有最大項(xiàng),無(wú)最小項(xiàng)C.無(wú)最大項(xiàng),有最小項(xiàng)D.無(wú)最大項(xiàng),無(wú)最小項(xiàng)答案B9.(2024上海,6,4分)記等差數(shù)列{an}的前n項(xiàng)和為Sn,若a3=0,a6+a7=14,則S7=.

答案1410.(2024課標(biāo)Ⅲ,14,5分)記Sn為等差數(shù)列{an}的前n項(xiàng)和,若a1≠0,a2=3a1,則S10S5答案411.(2024屆河南名校聯(lián)盟11月月考,15)已知等差數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,若SnTn=3n-12答案10綜合篇知能轉(zhuǎn)換考法一等差數(shù)列的判定1.(2024屆河南聯(lián)考三,9)在數(shù)列{an}中,2an-1·an+1=an-1an+anan+1(n≥2),a1=1,a2=13,則a20=(A.137B.139C.141答案B2.(2024屆安徽淮南一中月考三,11)已知數(shù)列{an}滿意a1=4,an=4an-1-4an-1(n≥2,n∈N*),若bn=412-an·(nan-6),且存在n∈N*,使得4bn+m-6mA.1-972,1+972C.0,16答案D3.(2024安徽馬鞍山一模,12)已知數(shù)列{an}滿意an+2+an=2an+1+1,且a1=1,a2=5,則a18=()A.69B.105C.204D.205答案D4.(2024甘肅頂級(jí)名校聯(lián)考,10)已知數(shù)列{an}的前n項(xiàng)和Sn=14n2+23n+3(n∈N*),則下列說(shuō)法正確的是(A.數(shù)列{an}是等差數(shù)列B.數(shù)列{an}是遞增數(shù)列C.a1,a5,a9成等差數(shù)列D.S6-S3,S9-S6,S12-S9成等差數(shù)列答案D5.(2024全國(guó)甲,18,12分)已知數(shù)列{an}的各項(xiàng)均為正數(shù),記Sn為{an}的前n項(xiàng)和,從下面①②③中選取兩個(gè)作為條件,證明另外一個(gè)成立.①數(shù)列{an}是等差數(shù)列;②數(shù)列{Sn}是等差數(shù)列;③a2=3a1注:若選擇不同的組合分別解答,則按第一個(gè)解答計(jì)分.解析選①②作為條件,證明③.證明:設(shè)等差數(shù)列{an}的公差為d,因?yàn)閧Sn}是等差數(shù)列,所以2S2=S1+S3,即22a1+d=a1+3a1+3d,兩邊平方,得4(2a1+d)=a1+3a1+3d+2a1(3a1+3d),整理得4a1+d=2a1(3a1+3d),兩邊平方,得16a12+8a1d+d2=4(3a1選①③作為條件,證明②.證明:設(shè)等差數(shù)列{an}的公差為d.因?yàn)閍2=3a1,即a1+d=3a1,所以d=2a1.所以等差數(shù)列{an}的前n項(xiàng)和Sn=na1+n(n-1)2d=na1+n(n-1)2·2a1=n2a1.又a1>0,所以Sn=na1.則Sn+1-Sn=(n+1)a選②③作為條件,證明①.證明:設(shè)等差數(shù)列{Sn}的公差為d,因?yàn)镾1=a1,S2=a1+a2=a1+3a1=2a1,所以d=S2-S1=2a1-a1=a1,則等差數(shù)列{Sn}的通項(xiàng)公式為Sn=a1+(n-1)a1=na1,所以Sn=n2a1,當(dāng)n≥2時(shí),an=Sn-Sn-1=n2a1-(n-1)2a1=(2n-1)a1,且當(dāng)n=1時(shí),上式也成立,所以數(shù)列{an}的通項(xiàng)公式為an=(2n-1)a方法總結(jié)證明數(shù)列{an}是等差數(shù)列的方法:(1)定義法:證明an-an-1=d(n≥2,n∈N*),或an+1-an=d(n∈N*);(2)等差中項(xiàng)法:證明2an=an-1+an+1(n≥2,n∈N*).6.(2024河南濮陽(yáng)模擬,17)已知數(shù)列{an}滿意an+1=1+an3-an(n∈N*),且(1)求證:數(shù)列1an-1是等差數(shù)列,并求(2)令bn=2(n+2)2an(n∈N*),求數(shù)列{bn解析(1)∵an+1=1+an3-an,∴an+1-1=1+an3-an-1=2an-23-an,故1a∴數(shù)列1an-1是公差為-又a1=13,∴1a1-1=113-1=-32,∴1an-1=-32-12(2)由(1)知an=nn+2,∴bn=2(n+2)2·nn+2=2n(n+2)=1n-1n+2,故Tn=b1+b2+…+bn=11-13+考法二等差數(shù)列前n項(xiàng)和的最值問(wèn)題1.(2024屆云南玉溪峨山一中月考,11)已知等差數(shù)列{an}的公差不等于0,其前n項(xiàng)和為Sn,若a4,S5,S7∈{-10,0},則Sn的最小值為()A.-6B.-11C.-12D.-14答案C2.(2024北京,10,5分)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn.若a2=-3,S5=-10,則a5=,Sn的最小值為.

答案0;-103.(2024青海西寧大通期末,17)已知前n項(xiàng)和為Sn的等差數(shù)列{an}的通項(xiàng)公式為an=40-3n.(1)求Sn的最大值;(2)令bn=1(an-40)(an+1-40),記數(shù)列{bn}的前n項(xiàng)和為T(mén)n,求滿意解析(1)解法一:令a