高考真題專題分類匯編等差數(shù)列（十年高考）含答案與解析

7.2等差數(shù)列考點(diǎn)1等差數(shù)列及其前n項(xiàng)和1.(2023全國(guó)甲文,5,5分,易)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a2+a6=10,a4a8=45,則S5=()A.25B.22C.20D.15答案C設(shè)等差數(shù)列{an}的公差為d,由已知可得2a1+6d=10,(a1+3d)(a1+7d)=45,2.(2023全國(guó)乙理,10,5分,中)已知等差數(shù)列{an}的公差為2π3,集合S={cosan|n∈N*}.若S={a,b},則ab=()A.-1B.-1答案B由題意,得an=a1+(n-1)·2π3,又S={cosan|n∈N*}={a,b},∴cosa1≠cosa2,但cosa1=cosa3,即cosa1=cosa1+4π3,∴a1+a1+4π3=2kπ(k∈Z),∴a1=-2π3+kπ(k∈Z).不妨取k=1,則a1=π3,a2=π3+2π3=π,則S=123.(2021北京,6,4分)已知{an}和{bn}是兩個(gè)等差數(shù)列,且akbk(1≤k≤5)是常數(shù),若a1=288,a5=96,b1=192,則b3的值為(A.64B.100C.128D.132答案C命題意圖:本題以定義新運(yùn)算為出題背景,考查等差數(shù)列的性質(zhì),考查學(xué)生的運(yùn)算求解能力,考查的核心素養(yǎng)是邏輯推理及數(shù)學(xué)運(yùn)算,落實(shí)了應(yīng)用性、綜合性和創(chuàng)新性的考查要求.解題思路:解法一:因?yàn)閧an}是等差數(shù)列,所以2a3=a1+a5=288+96,所以a3=192,又因?yàn)楫?dāng)1≤k≤5時(shí),akbk是常值,所以a3b3=a1解法二:由題意可知a1b1=a5b5,則b5=64,又{bn}是等差數(shù)列4.(2022新高考Ⅱ,3,5分)圖1是中國(guó)古代建筑中的舉架結(jié)構(gòu),AA',BB',CC',DD'是桁,相鄰桁的水平距離稱為步,垂直距離稱為舉.圖2是某古代建筑屋頂截面的示意圖,其中DD1,CC1,BB1,AA1是舉,OD1,DC1,CB1,BA1是相等的步,相鄰桁的舉步之比分別為DD1OD1=0.5,CC1DC1=k1,BB1CB1=k2,AA1BA1=k3.已知k1,k2,k3成公差為圖1圖2A.0.75B.0.8C.0.85D.0.9答案D令OD1=DC1=CB1=BA1=a,則CC1=k1a,BB1=k2a,AA1=k3a,DD1=0.5a,所以A的坐標(biāo)為(4a,k1a+k2a+k3a+0.5a),所以kOA=k1a+k所以k1+k2+k3+0.5=2.9.因?yàn)閗1,k2,k3成公差為0.1的等差數(shù)列,所以k1+k2+k3+0.5=3k3-0.3+0.5=2.9,所以k3=0.9,故選D.5.(2023新課標(biāo)Ⅰ,7,5分,中)記Sn為數(shù)列{an}的前n項(xiàng)和,設(shè)甲:{an}為等差數(shù)列;乙:Snn為等差數(shù)列,則(A.甲是乙的充分條件但不是必要條件B.甲是乙的必要條件但不是充分條件C.甲是乙的充要條件D.甲既不是乙的充分條件也不是乙的必要條件答案C若{an}為等差數(shù)列,設(shè)公差為d,則an=a1+(n-1)d,∴Sn=na1+n(n?1)d2,∴Snn當(dāng)n≥2時(shí),Sn?1n?1=a1∴Snn-Sn?1n?1=a1+n?12d-a∴Snn是以S1為首項(xiàng),d若Snn為等差數(shù)列,設(shè)公差為d',則Snn=S1+(n-1)d'=a1+(n∴Sn=na1+n(n-1)d',當(dāng)n≥2時(shí),Sn-1=(n-1)a1+(n-1)(n-2)d',兩式作差得,an=a1+2(n-1)d',又n=1時(shí)也滿足上式,∴an=a1+2(n-1)d',n∈N*,當(dāng)n≥2時(shí),an-1=a1+2(n-2)d',∴an-an-1=a1+2(n-1)d'-a1-2(n-2)d'=2d',∴{an}是以a1為首項(xiàng),2d'為公差的等差數(shù)列.綜上,甲是乙的充要條件,故選C.6.(2020課標(biāo)Ⅱ理,4,5分)北京天壇的圜丘壇為古代祭天的場(chǎng)所,分上、中、下三層.上層中心有一塊圓形石板(稱為天心石),環(huán)繞天心石砌9塊扇面形石板構(gòu)成第一環(huán),向外每環(huán)依次增加9塊.下一層的第一環(huán)比上一層的最后一環(huán)多9塊,向外每環(huán)依次也增加9塊.已知每層環(huán)數(shù)相同,且下層比中層多729塊,則三層共有扇面形石板(不含天心石)()A.3699塊B.3474塊C.3402塊D.3339塊答案C由題意可設(shè)每層有n個(gè)環(huán),則三層共有3n個(gè)環(huán),∴每一環(huán)扇面形石板的塊數(shù)構(gòu)成以a1=9為首項(xiàng),9為公差的等差數(shù)列{an},且項(xiàng)數(shù)為3n.不妨設(shè)上層扇面形石板總數(shù)為S1,中層總數(shù)為S2,下層總數(shù)為S3,∴S3-S2=9(2n+1)·n+n(n?1)2×9?9(則三層共有扇面形石板(不含天心石)27×9+27×262×9=27×9+27×13×9=27×14×9=3402(塊).故選C7.(2015課標(biāo)Ⅱ文,5,5分)設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和.若a1+a3+a5=3,則S5=()A.5B.7C.9D.11答案A∵{an}為等差數(shù)列,∴a1+a5=2a3,得3a3=3,則a3=1,∴S5=5(a1+a58.(2015課標(biāo)Ⅰ文,7,5分)已知{an}是公差為1的等差數(shù)列,Sn為{an}的前n項(xiàng)和.若S8=4S4,則a10=()A.172B.19答案B由S8=4S4得8a1+8×72×1=4×4a1+4×32×1,解得a評(píng)析本題主要考查等差數(shù)列的前n項(xiàng)和,計(jì)算準(zhǔn)確是解題關(guān)鍵,屬容易題.9.(2015浙江理,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B由a42=a3a8,得(a1+2d)(a1+7d)=(a1+3d)2,整理得d(5d+3a1)=0,又d≠0,∴a1=-53d,則a1d=-53d2<0,又∵S4=4a1+6d=-23d,∴dS4=-210.(2013課標(biāo)Ⅰ理,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C解法一:∵Sm-1=-2,Sm=0,Sm+1=3,∴am=Sm-Sm-1=2,am+1=Sm+1-Sm=3,∴公差d=am+1-am=1,由Sn=na1+n(n?1)2得m由①得a1=1?m2,代入解法二:∵數(shù)列{an}為等差數(shù)列,且前n項(xiàng)和為Sn,∴數(shù)列Snn∴Sm?1m?1+Sm+1m+1=2Sm11.(2016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C設(shè){an}的公差為d,由等差數(shù)列前n項(xiàng)和公式及通項(xiàng)公式,得S9=9a1+9×82d=27,思路分析用a1,d表示S9,a10,列方程組求出a1,d,從而可求得a100.12.(2022全國(guó)乙文,13,5分)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若2S3=3S2+6,則公差d=.

答案2解析∵2S3=3S2+6,∴2(3a1+3d)=3(2a1+d)+6,即6d=3d+6,解得d=2.13.(2014天津理,11,5分)設(shè){an}是首項(xiàng)為a1,公差為-1的等差數(shù)列,Sn為其前n項(xiàng)和.若S1,S2,S4成等比數(shù)列,則a1的值為.

答案-1解析S1=a1,S2=2a1-1,S4=4a1-6.故(2a1-1)2=a1×(4a1-6),解得a1=-1214.(2013課標(biāo)Ⅱ理,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解析由Sn=na1+n(n?解得a1=-3,d=23則Sn=-3n+n(n?1)2·2所以nSn=13(n3-10n2令f(x)=13(x3-10x2則f'(x)=x2-203x=xx?203,當(dāng)x∈1,203當(dāng)x∈203,+∞時(shí),f(x)遞增,又6<20f(7)=-49,所以nSn的最小值為-49.評(píng)析本題考查了數(shù)列與函數(shù)的應(yīng)用,考查了數(shù)列的基本運(yùn)算,利用導(dǎo)數(shù)求最值.本題易忽略n的取值范圍.15.(2023全國(guó)乙文,18,12分,中)記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a2=11,S10=40.(1)求{an}的通項(xiàng)公式;(2)求數(shù)列{|an|}的前n項(xiàng)和Tn.解析(1)設(shè)等差數(shù)列{an}的公差為d,則a1+∴an=13-2(n-1)=15-2n.(2)由an=15-2n知,當(dāng)n≤7,n∈N*時(shí),an>0,當(dāng)n≥8,n∈N*時(shí),an<0,∴當(dāng)n≤7,n∈N*時(shí),Tn=|a1|+|a2|+…+|an|=a1+a2+…+an=Sn=n(28?2n)2=-n當(dāng)n≥8,n∈N*時(shí),Tn=(a1+a2+…+a7)-(a8+a9+…+an)=2S7-Sn=98-(-n2+14n)=n2-14n+98.∴Tn=?16.(2023新課標(biāo)Ⅰ,20,12分,中)設(shè)等差數(shù)列{an}的公差為d,且d>1,令bn=n2+nan,記Sn,Tn分別為數(shù)列{an},{bn(1)若3a2=3a1+a3,S3+T3=21,求{an}的通項(xiàng)公式;(2)若{bn}為等差數(shù)列,且S99-T99=99,求d.解析(1)∵3a2=3a1+a3,∴3(a1+d)=3a1+a1+2d,∴a1=d>1,∴S3=a1+a2+a3=a1+a1+d+a1+2d=6a1,又∵bn=n2+nan,∴b1=2a1,b2=6a2=6a1+d=3a1,b3=12a3=12a∴S3+T3=6a1+9a1解得a1=3或a1=12(舍),∴an=3(2)∵{bn}為等差數(shù)列,∴2b2=b1+b3,即12a2=2a即6a1+d=1a1+6a1+2d,即∴a1=2d或a1=d.當(dāng)a1=2d時(shí),an=(n+1)d,bn=nd∴S99=(2d+100dT99=1d·(1+99)×992=又∵S99-T99=99,∴99×51d-99×50·1d=99∴51d-50d=1,解得d=1或d=-50又∵d>1,∴a1≠2d.當(dāng)a1=d時(shí),an=nd,bn=n+1∴S99=(1+99)×99d2=50×99T99=1d·(2+100)×992=又∵S99-T99=99,∴50×99d-51×99d=99∴50d-51d=1,解得d=5150或d又∵d>1,∴d=5150綜上,d=515017.(2022全國(guó)甲,理17,文18,12分)記Sn為數(shù)列{an}的前n項(xiàng)和.已知2Snn+n=2a(1)證明:{an}是等差數(shù)列;(2)若a4,a7,a9成等比數(shù)列,求Sn的最小值.解析(1)證明:由已知條件2Snn+n=2an2Sn=2nan+n-n2①,當(dāng)n≥2時(shí),由①可得2Sn-1=2(n-1)·an-1+(n-1)-(n-1)2②,由an=Sn-Sn-1及①-②可得,(2n-2)an-(2n-2)an-1=2n-2,n≥2,且n∈N*,即an-an-1=1,因此{(lán)an}是等差數(shù)列,公差為1.(2)∵a4,a7,a9成等比數(shù)列,且a4=a1+3,a7=a1+6,a9=a1+8,∴a72=a4·a9,即(a1+6)2=(a1+3)(a1解得a1=-12,∴等差數(shù)列{an}的通項(xiàng)公式為an=-12+(n-1)·1=n-13,∴Sn=(a∵n∈N*,∴當(dāng)n=12或n=13時(shí),Sn最小,最小值為-78.18.(2021全國(guó)乙理,19,12分)記Sn為數(shù)列{an}的前n項(xiàng)和,bn為數(shù)列{Sn}的前n項(xiàng)積,已知2Sn(1)證明:數(shù)列{bn}是等差數(shù)列;(2)求{an}的通項(xiàng)公式.解析(1)證明:由bn=S1·S2·…·Sn可得,Sn=b由2Sn+當(dāng)n=1時(shí),2S1+1b1=2,即2b1+1b當(dāng)n≥2時(shí),2bnbn?1+1bn=2,即2bn=2bn-1+1,即故數(shù)列{bn}是首項(xiàng)為32,公差為12(2)由(1)知,bn=32+(n-1)×1故當(dāng)n≥2時(shí),Sn=bnbn?1=即Sn=n+2n+1(n∈從而a1=S1=32當(dāng)n≥2時(shí),an=Sn-Sn-1=n+2n+1?n所以an=319.(2022浙江,20,15分)已知等差數(shù)列{an}的首項(xiàng)a1=-1,公差d>1.記{an}的前n項(xiàng)和為Sn(n∈N*).(1)若S4-2a2a3+6=0,求Sn;(2)若對(duì)于每個(gè)n∈N*,存在實(shí)數(shù)cn,使an+cn,an+1+4cn,an+2+15cn成等比數(shù)列,求d的取值范圍.解析(1)∵{an}為等差數(shù)列,∴an=(n-1)d-1,∴S4-2a2a3+6=6d-4-2(d-1)(2d-1)+6=0,解得d=3或d=0,又d>1,∴d=3,則an=3n-4,n∈N*,Sn=(?1+3n?4)n2=n(2)∵an+cn,an+1+4cn,an+2+15cn成等比數(shù)列,∴(an+1+4cn)2=(an+cn)(an+2+15cn),∴an+12+16cn2+8an+1cn=anan+2+15cn2+(∴cn2+(8an+1-an+2-15an)cn+d2由(1)知an=(n-1)d-1,∴8an+1-an+2-15an=8+(14-8n)d,則有cn2+[8+(14-8n)d]cn+d2=0對(duì)每個(gè)n∈N*∴Δ=[8+(14-8n)d]2-4d2=[8+(16-8n)d][8+(12-8n)d]≥0對(duì)每個(gè)n∈N*都成立,即n?1d?2n?1d由于n為正整數(shù),d>1,∴1d+2和1d+32只能同處于相鄰的整數(shù)之間,又∵2<1d+2<3∴23<d≤2,故1<d≤220.(2021全國(guó)甲文,18,12分)記Sn為數(shù)列{an}的前n項(xiàng)和,已知an>0,a2=3a1,且數(shù)列{Sn}是等差數(shù)列.證明:{an}是等差數(shù)列解題指導(dǎo):根據(jù)已知條件{Sn}為等差數(shù)列,求出數(shù)列{Sn}的通項(xiàng)公式,利用an=Sn-Sn-1(n≥2,n∈N*)即可求出{an}的通項(xiàng)公式,根據(jù)等差數(shù)列的定義可證明其為等差數(shù)列解析設(shè)等差數(shù)列{Sn}的公差為d,因?yàn)閍2=3a1,S1=a1,S2=a1+a2=4a1=2a1,所以d當(dāng)n≥2時(shí),Sn-1=(n-1)2a1,所以an=Sn-Sn-1=n2a1-(n-1)2a1=(2n-1)a1,經(jīng)檢驗(yàn),當(dāng)n=1時(shí),也滿足題意,所以an=(2n-1)a1,n∈N*,當(dāng)n≥2時(shí),an-an-1=(2n-1)a1-(2n-3)a1=2a1(常數(shù)),∴{an}是等差數(shù)列.易錯(cuò)警示應(yīng)用an=S1(n=1),Sn?21.(2021全國(guó)甲理,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.解題指導(dǎo):先選兩個(gè)作為條件,余下一個(gè)作為結(jié)論,然后進(jìn)行證明.如果是證明③,利用等差中項(xiàng)法,即只需2S2=S1+S3,通過化簡(jiǎn)即可得證;如果是證明①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(3a12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則Sn+1?Sn=(n+1)a1?na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=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*).22.(2013課標(biāo)Ⅱ文,17,12分)已知等差數(shù)列{an}的公差不為零,a1=25,且a1,a11,a13成等比數(shù)列.(1)求{an}的通項(xiàng)公式;(2)求a1+a4+a7+…+a3n-2.解析(1)設(shè){an}的公差為d.由題意,得a112=a1a即(a1+10d)2=a1(a1+12d).于是d(2a1+25d)=0.又a1=25,所以d=0(舍去)或d=-2.故an=-2n+27.(2)令Sn=a1+a4+a7+…+a3n-2.由(1)知a3n-2=-6n+31,故{a3n-2}是首項(xiàng)為25,公差為-6的等差數(shù)列.從而Sn=n2(a1+a3n-2=n2=-3n2+28n.23.(2014大綱全國(guó)文,17,10分)數(shù)列{an}滿足a1=1,a2=2,an+2=2an+1-an+2.(1)設(shè)bn=an+1-an,證明{bn}是等差數(shù)列;(2)求{an}的通項(xiàng)公式.解析(1)證明:由an+2=2an+1-an+2得,an+2-an+1=an+1-an+2,即bn+1=bn+2.又b1=a2-a1=1,所以{bn}是首項(xiàng)為1,公差為2的等差數(shù)列.(5分)(2)由(1)得bn=1+2(n-1),即an+1-an=2n-1.(8分)于是∑所以an+1-a1=n2,即an+1=n2+a1.又a1=1,所以{an}的通項(xiàng)公式為an=n2-2n+2.(10分)評(píng)析本題著重考查等差數(shù)列的定義、前n項(xiàng)和公式及“累加法”求數(shù)列的通項(xiàng)等基礎(chǔ)知識(shí),同時(shí)考查運(yùn)算變形的能力.24.(2014課標(biāo)Ⅰ理,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)