浙江專用2025屆高考數(shù)學(xué)一輪復(fù)習(xí)專題六數(shù)列6.2等差數(shù)列試題含解析

PAGEPAGE1§6.2等差數(shù)列基礎(chǔ)篇固本夯基【基礎(chǔ)集訓(xùn)】考點(diǎn)一等差數(shù)列的有關(guān)概念及運(yùn)算1.已知等差數(shù)列{an}中,a2=1,前5項(xiàng)和S5=-15,則數(shù)列{an}的公差為()A.-3B.-52答案D2.已知在等差數(shù)列{an}中,a1=1,a3=2a+1,a5=3a+2,若Sn=a1+a2+…+an,且Sk=66,則k的值為()A.9B.11C.10D.12答案B3.設(shè)等差數(shù)列{an}滿意3a8=5a15,且a1>0,Sn為其前n項(xiàng)和,則數(shù)列{Sn}的最大項(xiàng)為()A.S23B.S24C.S25D.S26答案C4.已知數(shù)列{an}滿意a1=12,且an+1=2(1)求證:數(shù)列1a(2)若bn=anan+1,求數(shù)列{bn}的前n項(xiàng)和Sn.解析(1)證明:易知an≠0,∵an+1=2a∴1an+1=2+an2a又∵a1=12,∴1∴數(shù)列1an是以2為首項(xiàng),(2)由(1)知,1an=2+12(n-1)=n+32∴bn=4(n+3∴Sn=41=414-1考點(diǎn)二等差數(shù)列的性質(zhì)5.設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,若a6a5=9A.1B.-1C.2D.1答案A6.(2024河北唐山其次次模擬,7)設(shè){an}是隨意等差數(shù)列,它的前n項(xiàng)和、前2n項(xiàng)和與前4n項(xiàng)和分別為X,Y,Z,則下列等式中恒成立的是()A.2X+Z=3YB.4X+Z=4YC.2X+3Z=7YD.8X+Z=6Y答案D7.已知數(shù)列{an}是公差為d的等差數(shù)列,Sn為其前n項(xiàng)和,若S2017A.120B.1答案B8.在等差數(shù)列{an}中,a3+a7=37,則a2+a4+a6+a8=.

答案749.已知An及Bn是等差數(shù)列{an}、{bn}的前n項(xiàng)和,且AnBn=3n+1答案6410.已知數(shù)列{an}是等差數(shù)列.(1)前四項(xiàng)和為21,末四項(xiàng)和為67,且各項(xiàng)和為286,求項(xiàng)數(shù);(2)項(xiàng)數(shù)為奇數(shù),奇數(shù)項(xiàng)和為44,偶數(shù)項(xiàng)和為33.求數(shù)列的中間項(xiàng)和項(xiàng)數(shù).解析(1)由已知得a1+a2+a3+a4=21,an-3+an-2+an-1+an=67,∴a1+a2+a3+a4+an-3+an-2+an-1+an=88,∴a1+an=884∵Sn=286,∴n((2)解法一:設(shè)項(xiàng)數(shù)為2k+1,則a1+a3+…+a2k+1=44=k+12(a1+a2k+1),a2+a4+…+a2k=33=k2(a2又∵a1+a2k+1=a2+a2k,∴k+1k=∴中間項(xiàng)為a1解法二:記等差數(shù)列{an}的中間項(xiàng)為a中,奇數(shù)項(xiàng)和為S奇,偶數(shù)項(xiàng)和為S偶,前n項(xiàng)和為Sn.依據(jù)題意得S偶+S奇=又na中=Sn,∴n=7.綜合篇知能轉(zhuǎn)換【綜合集訓(xùn)】考法一等差數(shù)列的判定與證明1.(2024山東濟(jì)寧一模,11)設(shè)數(shù)列{an}滿意a1=1,a2=2,且2nan=(n-1)an-1+(n+1)an+1(n≥2且n∈N*),則a18=()A.259B.269答案B2.(2024河北冀州模擬,9)已知{an},{bn}均為等差數(shù)列,且a2=4,a4=6,b3=3,b7=9,由{an},{bn}的公共項(xiàng)組成新數(shù)列{cn},則c10=()A.18B.24C.30D.36答案C3.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2n-1.數(shù)列{bn}滿意b1=2,bn+1-2bn=8an.(1)求數(shù)列{an}的通項(xiàng)公式;(2)證明數(shù)列bn2n解析(1)當(dāng)n=1時(shí),a1=S1=21-1=1;當(dāng)n≥2時(shí),an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1.因?yàn)閍1=1適合上式,所以an=2n-1(n∈N*).(2)因?yàn)閎n+1-2bn=8an,所以bn+1-2bn=2n+2,即bn+12n+1-bn2n=2.又b1考法二等差數(shù)列前n項(xiàng)和的最值問題4.(2024江西贛中南五校聯(lián)考,4)在等差數(shù)列{an}中,已知a3+a8>0,且S9<0,則S1、S2、…、S9中最小的是()A.S5B.S6C.S7D.S8答案A5.(2024廣東汕頭模擬,8)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a1=9,S99-S5A.4B.5C.6D.4或5答案B6.(2024湖南永州三模,11)已知數(shù)列{an}是等差數(shù)列,前n項(xiàng)和為Sn,滿意a1+5a3=S8,給出下列結(jié)論:①a10=0;②S10最小;③S7=S12;④S20=0.其中肯定正確的結(jié)論是()A.①②B.①③④C.①③D.①②④答案C7.(2024廣東深圳期末,14)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a1=-11,a4+a6=-6,則當(dāng)Sn取最小值時(shí),n=.

答案6【五年高考】考點(diǎn)一等差數(shù)列的有關(guān)概念及運(yùn)算1.(2024課標(biāo)Ⅰ,3,5分)已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=8,則a100=()A.100B.99C.98D.97答案C2.(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C4.(2024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A5.(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A6.(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47.(2024北京,9,5分)設(shè){an}是等差數(shù)列,且a1=3,a2+a5=36,則{an}的通項(xiàng)公式為.

答案an=6n-38.(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169.(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;-1010.(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.所以{an}的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.11.(2024天津,18,13分)已知{an}是各項(xiàng)均為正數(shù)的等差數(shù)列,公差為d.對(duì)隨意的n∈N*,bn是an和an+1的等比中項(xiàng).(1)設(shè)cn=bn+12-bn2(2)設(shè)a1=d,Tn=∑k=12n(-1)kbk2,n∈N證明(1)由題意得bn2=anan+1,有cn=bn+12-bn2=an+1·an+2-anan+1=2dan+1,因此cn+1-cn所以{cn}是等差數(shù)列.(2)Tn=(-b12+b22)+(-b32+=2d(a2+a4+…+a2n)=2d·n(a2所以∑k=1n1Tk=12d2考點(diǎn)二等差數(shù)列的性質(zhì)12.(2024廣東,10,5分)在等差數(shù)列{an}中,若a3+a4+a5+a6+a7=25,則a2+a8=.

答案10老師專用題組考點(diǎn)一等差數(shù)列的有關(guān)概念及運(yùn)算1.(2024浙江,6,5分)如圖,點(diǎn)列{An},{Bn}分別在某銳角的兩邊上,且|AnAn+1|=|An+1An+2|,An≠An+2,n∈N*,|BnBn+1|=|Bn+1Bn+2|,Bn≠Bn+2,n∈N*.(P≠Q(mào)表示點(diǎn)P與Q不重合)若dn=|AnBn|,Sn為△AnBnBn+1的面積,則()A.{Sn}是等差數(shù)列B.{SnC.{dn}是等差數(shù)列D.{dn答案A2.(2024浙江,3,5分)已知{an}是等差數(shù)列,公差d不為零,前n項(xiàng)和是Sn.若a3,a4,a8成等比數(shù)列,則()A.a1d>0,dS4>0B.a1d<0,dS4<0C.a1d>0,dS4<0D.a1d<0,dS4>0答案B3.(2013課標(biāo)Ⅰ,7,5分)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,則m=()A.3B.4C.5D.6答案C4.(2024江蘇,8,5分)已知{an}是等差數(shù)列,Sn是其前n項(xiàng)和.若a1+a22=-3,S5=10,則a9的值是答案205.(2024課標(biāo)Ⅰ,17,12分)已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,an≠0,anan+1=λSn-1,其中λ為常數(shù),(1)證明:an+2-an=λ;(2)是否存在λ,使得{an}為等差數(shù)列?并說明理由.解析(1)證明:由題設(shè)anan+1=λSn-1,知an+1an+2=λSn+1-1.兩式相減得,an+1(an+2-an)=λan+1.由于an+1≠0,所以an+2-an=λ.(2)存在.由a1=1,a1a2=λa1-1,可得a2=λ-1,由(1)知,a3=λ+1.令2a2=a1+a3,解得λ=4.故an+2-an=4,由此可得,{a2n-1}是首項(xiàng)為1,公差為4的等差數(shù)列,a2n-1=1+(n-1)·4=4n-3;{a2n}是首項(xiàng)為3,公差為4的等差數(shù)列,a2n=3+(n-1)·4=4n-1.所以an=2n-1,an+1-an=2.因此存在λ=4,使得{an}為等差數(shù)列.思路分析(1)已知anan+1=λSn-1,用n+1代替n得an+1·an+2=λSn+1-1,兩式相減得結(jié)論.(2)利用a1=1,a2=λ-1,a3=λ+1及2a2=a1+a3,得λ=4.進(jìn)而得an+2-an=4.故數(shù)列{an}的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)分別組成公差為4的等差數(shù)列,分別求通項(xiàng)公式,進(jìn)而求出{an}的通項(xiàng)公式,從而證出等差數(shù)列.方法總結(jié)對(duì)于含an、Sn的等式的處理,往往可轉(zhuǎn)換為關(guān)于an的遞推式或關(guān)于Sn的遞推式;對(duì)于存在性問題,可先探求參數(shù)的值再證明.考點(diǎn)二等差數(shù)列的性質(zhì)6.(2024陜西,13,5分)中位數(shù)為1010的一組數(shù)構(gòu)成等差數(shù)列,其末項(xiàng)為2015,則該數(shù)列的首項(xiàng)為.

答案57.(2013課標(biāo)Ⅱ,16,5分)等差數(shù)列{an}的前n項(xiàng)和為Sn.已知S10=0,S15=25,則nSn的最小值為.

答案-49【三年模擬】一、單項(xiàng)選擇題(每題5分,共40分)1.(2025屆云南陸良其次次教學(xué)質(zhì)量摸底考,3)已知{an}為等差數(shù)列,若a3+a4+a8=12,則S9=()A.24B.27C.36D.54答案C2.(2025屆四川宜賓四中開學(xué)考,4)已知等差數(shù)列{an}中,a2、a2016是方程x2-2x-2=0的兩根,則S2017=()A.-2017B.-1008C.1008D.2017答案D3.(2025屆河北邯鄲大名一中第六周周測(cè),4)設(shè){an}是等差數(shù)列,則下列結(jié)論肯定正確的是()A.若a1+a2>0,則a2+a3>0B.若a1+a3<0,則a2+a3<0C.若0<a1<a2,則a2>aD.(a2-a1)(a2-a3)<0答案C4.(20245·3原創(chuàng)沖刺卷一,4)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S2=3,S3=6,則S2n+1=()A.(2n+1)(n+1)B.(2n+1)(n-1)C.(2n-1)(n+1)D.(2n+1)(n+2)答案A5.(2024安徽合肥其次次教學(xué)質(zhì)量檢測(cè),5)中國(guó)古詩(shī)詞中,有一道“八子分綿”的數(shù)學(xué)名題:“九百九十六斤綿,贈(zèng)分八子作盤纏,次第每人多十七,要將第八數(shù)來言”.題意是:把996斤綿分給8個(gè)兒子作盤纏,依據(jù)年齡從大到小的依次依次分綿,年齡小的比年齡大的多17斤綿,那么第8個(gè)兒子分到的綿是()A.174斤B.184斤C.191斤D.201斤答案B6.(2024湖北宜昌一模,8)等差數(shù)列{an}的前n項(xiàng)和為Sn,若公差d>0,(S8-S5)(S9-S5)<0,則()A.a7=0B.|a7|=|a8|C.|a7|>|a8|D.|a7|<|a8|答案D7.(2024湖南三湘名校教化聯(lián)盟第三次聯(lián)考,5)已知等差數(shù)列{an}的各項(xiàng)都為整數(shù),且a1=-5,a3a4=-1,則|a1|+|a2|+…+|a10|=()A.70B.58C.51D.40答案B8.(2024安徽淮北一模,9)Sn是等差數(shù)列{an}的前n項(xiàng)和,S2018<S2016,S2017<S2018,則Sn<0時(shí)n的最大值是()A.2017B.2018C.4033D.4034答案D二、多項(xiàng)選擇題(每題5分,共10分)9.(改編題)設(shè)等差數(shù)列{an}滿意a3+a7=36,a4a6=275,則()A.an=7n-17B.an=-7n+53C.anan+1的最小值為-12D.anan+1無最小值答案ABC10.(改編題)記Sn為等比數(shù)列{an}的前n項(xiàng)和,已知S2=2,S3=-6,則()A.an=(-2)nB.Sn+1,Sn,Sn+2成等差數(shù)列C.an=-2nD.Sn+1,Sn,Sn+2不成等差數(shù)列答案AB三、填空題(每題5分,共15分)11.(2025屆河北邯鄲大名一中周測(cè),14)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a2=3,S4=16,則數(shù)列{an}的公差d=.

答案212.(2024上海嘉定(長(zhǎng)寧)二模,11)已知有窮數(shù)列{an}共有m項(xiàng),記數(shù)列{an}的全部項(xiàng)的和為S(1),其次項(xiàng)及以后全部項(xiàng)的和為S(