【高考數(shù)學(xué)】2026版53高考總復(fù)習(xí)A版數(shù)學(xué)61數(shù)列的概念及表示專題六數(shù)列6.1數(shù)列的概念及表示五年高考考點(diǎn)數(shù)列的概念及表示1.(2022全國(guó)乙理,4,5分,中)嫦娥二號(hào)衛(wèi)星在完成探月任務(wù)后,繼續(xù)進(jìn)行深空探測(cè),成為我國(guó)第一顆環(huán)繞太陽飛行的人造行星.為研究嫦娥二號(hào)繞日周期與地球繞日周期的比值,用到數(shù)列{bn}:b1=1+1α1,b2=1+1α1+1α2,b3=1+1α1+1α2+1α3,…,依此類推,其中αA.b1<b5B.b3<b8C.b6<b2D.b4<b7答案D2.(2021浙江,10,5分,難)已知數(shù)列{an}滿足a1=1,an+1=an1+an(n∈N*).記數(shù)列{an}的前n項(xiàng)和為Sn,則A.32<S100<3B.3<S100C.4<S100<92D.92答案A3.(2019上海,8,5分,易)已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足Sn+an=2,則S5=.

答案314.(2014課標(biāo)Ⅱ,16,5分,易)數(shù)列{an}滿足an+1=11?an,a8=2,則a1答案15.(2015江蘇,11,5分,中)設(shè)數(shù)列{an}滿足a1=1,且an+1-an=n+1(n∈N*),則數(shù)列1an前10項(xiàng)的和為答案206.(2022北京,15,5分,難)已知數(shù)列{an}的各項(xiàng)均為正數(shù),其前n項(xiàng)和Sn滿足an·Sn=9(n=1,2,…).給出下列四個(gè)結(jié)論:①{an}的第2項(xiàng)小于3;②{an}為等比數(shù)列;③{an}為遞減數(shù)列;④{an}中存在小于1100的項(xiàng)其中所有正確結(jié)論的序號(hào)是.

答案①③④7.(2016課標(biāo)Ⅲ,17,12分,中)已知各項(xiàng)都為正數(shù)的數(shù)列{an}滿足a1=1,an2-(2an+1-1)an-2an+1(1)求a2,a3;(2)求{an}的通項(xiàng)公式.解析(1)令n=1,則由題可得a12-(2a2-1)a1-2a2把a(bǔ)1=1代入,得1-(2a2-1)-2a2=0,解得a2=12令n=2,則a22-(2a3-1)a2-2a3即14?12(2a3-1)-2a3=0,解得a(2)將an2-(2an+1-1)an-2an+1=0因式分解可得(an+1)·(an-2an+1)=0,則an=-1(舍)或an-2an+1即an+1an=12,即{an}是以1為首項(xiàng),12三年模擬基礎(chǔ)強(qiáng)化練1.(2025屆安徽合肥一六八中學(xué)月考,3)已知數(shù)列{an}滿足an+1(1-an)=1,若a1=-1,則a10=()A.2B.-2C.-1D.1答案C2.(2025屆廣東汕頭摸底考,5)已知數(shù)列an=n?2025n?2024(n∈N*),則數(shù)列{an}A.a1,a100B.a45,a44C.a45,a1D.a44,a100答案B3.(2025屆遼寧沈陽聯(lián)考,6)已知數(shù)列{an}的前n項(xiàng)和為Sn.若an+an+1=2n+5,a1=1,則S8=()A.48B.50C.52D.54答案C4.(2025屆重慶西南大學(xué)附中階段性檢測(cè),7)已知數(shù)列{an}的首項(xiàng)a1=2025,其前n項(xiàng)和Sn滿足Sn=n2an,則a2024=()A.1答案C5.(2025屆浙江名校聯(lián)考,6)數(shù)列{an}滿足an+2=2an+1+3an,則下列a1,a2的值能使數(shù)列{an}為周期數(shù)列的是()A.a1=0,a2=1B.a1=-1,a2=1C.a1=0,a2=2D.a1=-2,a2=0答案B6.(2025屆遼寧沈陽東北育才學(xué)校開學(xué)考,12)若數(shù)列{an}滿足a1=1,an+1-an=n+1,數(shù)列1an的前n項(xiàng)和為Sn,則S2024=答案40487.(2025屆安徽江淮十校聯(lián)考,13)記Sn為數(shù)列{an}的前n項(xiàng)和.已知3Snn+n=3an+1,a1=23,則數(shù)列{an答案an=238.(2024浙江寧波模擬,18)已知數(shù)列{an}滿足a1=1,且對(duì)任意正整數(shù)m,n都有am+n=an+am+2mn.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求數(shù)列{(-1)nan}的前n項(xiàng)和Sn.解析(1)由對(duì)任意正整數(shù)m,n均有am+n=an+am+2mn,取m=1,得an+1=an+1+2n,當(dāng)n≥2時(shí),an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+3+5+…+2n-1=n(1+2n?1)2當(dāng)n=1時(shí),a1=1,符合上式,所以an=n2,n∈N*.(2)當(dāng)n為偶數(shù)時(shí),Sn=(-12+22)+(-32+42)+…+[-(n-1)2+n2]=3+7+11+…+(2n-1)=n2當(dāng)n為奇數(shù)時(shí),Sn=Sn-1+(-1)nan=Sn-1-an=(n綜上所述:Sn=n

能力拔高練1.(2025屆福建寧德福鼎一中月考,8)已知正項(xiàng)數(shù)列{bn}滿足b1=1,bn=nbn+12nA.b2=1+52B.{bnC.bn+1-bn<1答案C2.(2024遼寧大連期末,8)已知函數(shù)f(x),若數(shù)列an=f(n),n∈N*為遞增數(shù)列,則稱函數(shù)f(x)為“數(shù)列保增函數(shù)”,已知函數(shù)f(x)=-ln(2x)+λx為“數(shù)列保增函數(shù)”,則λ的取值范圍是()A.ln32,+∞B.(C.[1,+∞)D.1答案B3.(2025屆廣東一調(diào),7)斐波那契數(shù)列因數(shù)學(xué)家斐波那契以兔子繁殖為例而引入,又稱“兔子數(shù)列”.這一數(shù)列如下定義:設(shè){an}為斐波那契數(shù)列,a1=1,a2=1,an=an-1+an-2(n≥3,n∈N*),其通項(xiàng)公式為an=151+52n?1?52n,設(shè)n是log2[(1+5)x-(1-5)x]<A.5B.6C.7D.8答案A4.(2025屆廣東深圳開學(xué)考,8)數(shù)列{an}中,a1=2,an+1=an2-an+1,記An=1a1+1a2+…+1aA.A2024+B2024>1B.A2024+B2024<1C.A2024-B2024>1答案C5.(多選)(2025屆山東青島五十八中期中,10)若數(shù)列{an}滿足a1=1,a2=1,an=an-1+an-2(n≥3,n∈N*),則稱數(shù)列{an}為斐波那契數(shù)列,又稱黃金分割數(shù)列,則下列結(jié)論成立的是()A.a7=13B.2an=an-2+an+2(n≥3,n∈N*)C.a1+a3+a5+…+a2023=a2024D.a2+a4+a6+…+a2024=a2025答案AC6.(2025屆云南名校聯(lián)考,13)自然常數(shù)e是自然對(duì)數(shù)的底數(shù),大約等于2.71828.某人用“調(diào)日法”找逼近e的分?jǐn)?shù),稱小于2.718281的值為弱值,大于2.718282的值為強(qiáng)值.由21<e<31,取2為弱值,3為強(qiáng)值,得a1=2+31+1=52,故a1為弱值,與上一次的強(qiáng)值3計(jì)算得a2=5+32+1=83,故a2為弱值,繼續(xù)計(jì)算,……,若某次得到的近似值為弱值,與上一次的強(qiáng)值繼續(xù)計(jì)算得到新的近似值;若某次得到的近似值為強(qiáng)值,與上一次的弱值繼續(xù)計(jì)算得到新的近似值,答案67.(2025屆河北邯鄲第一次調(diào)研,14)已知有窮遞增數(shù)列{an}的各項(xiàng)均為正整數(shù)(n≥3),所有項(xiàng)的和為S,所有項(xiàng)的積為T,若T=4S,則該數(shù)列可能為.(填寫一個(gè)數(shù)列即可)

答案1,5,24(答案不唯一)8.(2025屆北京清華大學(xué)附屬中學(xué)開學(xué)考,15)已知數(shù)列{an}滿足a1=a,an+1=12an2+2(n=1,2,①當(dāng)a=-1時(shí),存在k∈N*,使得ak=2;②當(dāng)a=1時(shí),{an}為遞增數(shù)列,且an<2恒成立;③存在a∈R,使得{an}中既有最大值,又有最小值;④對(duì)任意的a∈R,存在n0∈N*,當(dāng)n>n0時(shí),|an-2|<12024恒成立其中,所有正確結(jié)論的序號(hào)為.

答案②③④6.2等差數(shù)列五年高考考點(diǎn)1等差數(shù)列及其前n項(xiàng)和1.(2024全國(guó)甲文,5,5分,易)記Sn為等差數(shù)列{an}的前n項(xiàng)和.已知S9=1,則a3+a7=()A.1答案B2.(2024全國(guó)甲理,4,5分,易)記Sn為等差數(shù)列{an}的前n項(xiàng)和.已知S5=S10,a5=1,則a1=()A.72C.-1答案B3.(2019課標(biāo)Ⅰ理,9,5分,中)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=12n2-2答案A4.(2023全國(guó)甲文,5,5分,中)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C5.(2022新高考Ⅱ,3,5分,中)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,A.0.75B.0.8C.0.85D.0.9答案D6.(2021北京,6,4分,中)《中國(guó)共產(chǎn)黨黨旗黨徽制作和使用的若干規(guī)定》指出,中國(guó)共產(chǎn)黨黨旗為旗面綴有金黃色黨徽?qǐng)D案的紅旗,通用規(guī)格有五種.這五種規(guī)格黨旗的長(zhǎng)a1,a2,a3,a4,a5(單位:cm)成等差數(shù)列,對(duì)應(yīng)的寬為b1,b2,b3,b4,b5(單位:cm),且長(zhǎng)與寬之比都相等.已知a1=288,a5=96,b1=192,則b3=()A.64B.96C.128D.160答案C7.(2024新課標(biāo)Ⅱ,12,5分,易)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a3+a4=7,3a2+a5=5,則S10=.

答案958.(2020課標(biāo)Ⅱ文,14,5分,易)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a1=-2,a2+a6=2,則S10=.

答案259.(2019北京理,10,5分,中)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.(2020新高考Ⅰ,14,5分,中)將數(shù)列{2n-1}與{3n-2}的公共項(xiàng)從小到大排列得到數(shù)列{an},則{an}的前n項(xiàng)和為.

答案3n2-2n11.(2019課標(biāo)Ⅰ文,18,12分,中)記Sn為等差數(shù)列{an}的前n項(xiàng)和.已知S9=-a5.(1)若a3=4,求{an}的通項(xiàng)公式;(2)若a1>0,求使得Sn≥an的n的取值范圍.解析(1)設(shè){an}的公差為d.由S9=-a5得a1+4d=0.由a3=4得a1+2d=4.于是a1=8,d=-2.因此{(lán)an}的通項(xiàng)公式為an=10-2n.(2)由(1)得a1=-4d,故an=(n-5)d,Sn=n(n?9)d2.由a1>0知d<0,故Sn≥an等價(jià)于n2-11n+10≤0,解得1≤n≤10.所以n的取值范圍是{n|1≤n≤10,n12.(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+d=11,10a1+45d=40,(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=?13.(2021全國(guó)乙理,19,12分,中)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=b1,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bn-bn-1=12,故數(shù)列{(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∈N*),從而a1=S1當(dāng)n≥2時(shí),an=Sn-Sn-1=n+2n+1?n所以an=314.(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,b∴S3+T3=6a1+9a1=21,解得a1=3或a1=12(舍),∴a(2)∵{bn}為等差數(shù)列,∴2b2=b1+b3,即12a即6a1+d=1a1+6a1∴a1=2d或a1=d.當(dāng)a1=2d時(shí),an=(n+1)d,bn=nd∴S99=(2d+100dT99=1d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+1d,∴S99=(1+99)×99dT99=1d又∵S99-T99=99,∴50×99d-51×99d=99∴50d-51d=1,解得d=5150或d=-1,又∵d>1,∴d=綜上,d=5150考點(diǎn)2等差數(shù)列的性質(zhì)1.(2020浙江,7,4分,中)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,公差d≠0,且a1d≤1.記b1=S2,bn+1=S2n+2-S2n,n∈N*,下列等式不可能成立的是(A.2a4=a2+a6B.2b4=b2+b6C.a42=a2答案D2.(2020北京,8,4分,難)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),無最小項(xiàng)C.無最大項(xiàng),有最小項(xiàng)D.無最大項(xiàng),無最小項(xiàng)答案B3.(2021新高考Ⅱ,17,10分,中)記Sn為公差不為零的等差數(shù)列{an}的前n項(xiàng)和,若a3=S5,a2·a4=S4.(1)求{an}的通項(xiàng)公式;(2)求使得Sn>an的n的最小值.解析(1)解法一:設(shè)等差數(shù)列{an}的公差為d(d≠0),則a3=S5?a1+2d=5a1+10d?4a1+8d=0?a1+2d=0?a1=-2d,①a2·a4=S4?(a1+d)(a1+3d)=4a1+6d,②將①代入②得-d2=-2d?d=0(舍)或d=2,∴a1=-2d=-4,∴an=-4+(n-1)×2=2n-6.解法二:由等差數(shù)列的性質(zhì)可得S5=5a3,則a3=5a3,∴a3=0,設(shè)等差數(shù)列的公差為d,從而有a2a4=(a3-d)(a3+d)=-d2,S4=a1+a2+a3+a4=(a3-2d)+(a3-d)+a3+(a3+d)=-2d,從而-d2=-2d,由于公差不為零,故d=2,所以數(shù)列{an}的通項(xiàng)公式為an=a3+(n-3)d=2n-6.(2)由(1)知an=2n-6,a1=2×1-6=-4.Sn=na1+n(n?1)2d=-4n+n(n-1)=Sn>an?n2-5n>2n-6?n2-7n+6>0?(n-1)(n-6)>0,解得n<1或n>6,又n∈N*,∴n的最小值為7.4.(2022浙江,20,15分,中)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)易得an=(n-1)d-1,n∈N*,S4=a1+a2+a3+a4=4a1+6d=6d-4.又S4-2a2a3+6=0,∴6d-4-2(d-1)(2d-1)+6=0,∴d=3或d=0(舍),則an=3n-4,n∈N*,故Sn=3(1+2+…+n)-4n=3n(n+1)?8n2(2)由(1)知an=(n-1)d-1,n∈N*,依題意得[cn+(n-1)d-1][15cn+(n+1)d-1]=(4cn+nd-1)2,即15cn2+[(16n-14)d-16]cn+(n2-1)d2-2nd+1=16cn2+8(nd-1)·cn+n2d2-2nd+1,故cn2+[(14-8n)d+8]c故[(14-8n)d+8]2-4d2=[(12-8n)d+8][(16-8n)d+8]≥0,故[(3-2n)d+2][(2-n)d+1]≥0對(duì)任意正整數(shù)n恒成立,n=1時(shí),顯然成立;n=2時(shí),-d+2≥0,則d≤2;n≥3時(shí),[(2n-3)d-2][(n-2)d-1]>(2n-5)(n-3)≥0.綜上所述,1<d≤2.三年模擬基礎(chǔ)強(qiáng)化練1.(2025屆江西多校聯(lián)考,3)已知等差數(shù)列{an}的公差d≠0,若a1=3,且a2,a4,a8成等比數(shù)列,則d=()A.2B.3C.5答案B2.(2025屆內(nèi)蒙古包頭聯(lián)考,3)“數(shù)列{an}是等差數(shù)列”是“數(shù)列{an+an+1}是等差數(shù)列”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件答案A3.(2025屆廣東三校摸底考試,5)在等差數(shù)列{an}中,S3=3,S6=10,則S9=()A.13B.17C.21D.23答案C4.(2025屆山東濰坊開學(xué)考,6)數(shù)列{an}中,a1=2,an+1=an+2,若ak+ak+1+…+ak+9=270,則k=()A.7B.8C.9D.10答案C5.(多選)(2025屆海南瓊海開學(xué)考,9)已知數(shù)列{an}的前n項(xiàng)和為Sn,若a1=-10,an+1=an+3,則下列說法正確的是()A.{an}是遞增數(shù)列B.10是數(shù)列{an}中的項(xiàng)C.數(shù)列SnD.數(shù)列{Sn}中的最小項(xiàng)為S3答案AC6.(2025屆湖南長(zhǎng)沙雅禮中學(xué)期中,16)已知公差不為零的等差數(shù)列{an}的前n項(xiàng)和為Sn,若S10=110,且a1,a2,a4成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=an+3an,求數(shù)列{bn}的前n解析(1)設(shè)公差為d,d≠0.由S10=110,得110=10a1+45d①,∵a1,a2,a4成等比數(shù)列,∴a22=a1·a4,∴(a1+d)2=a1(a1+3d)由①②解得a∴數(shù)列{an}的通項(xiàng)公式為an=2n.(2)由(1)得bn=an+3an=2n+32n=2n+9則數(shù)列{bn}的前n項(xiàng)和為b1+b2+…+bn=2×(1+2+…+n)+(9+92+…+9n)=2×n(n+1)2+9(1?9n)7.(2025屆河北石家莊重點(diǎn)高中開學(xué)考,15)已知數(shù)列{an}是等差數(shù)列,且a1=1,a3=a2+2a1,Sn是數(shù)列{an}的前n項(xiàng)和.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=1SnSn+1(n∈N*),數(shù)列{bn}的前n項(xiàng)和為Tn,求證解析(1)設(shè)公差為d.由a3=a2+2a1,得a1+2d=a1+d+2a1,則d=2a1=2,因此an=a1+(n-1)d=2n-1,故數(shù)列{an}的通項(xiàng)公式為an=2n-1.(2)證明:由(1)知,an=2n-1,則Sn=n(a1+則1SnSn+1=1則Tn=1?1由于n∈N*,則1n+1>0,故Tn8.(2025屆湖北宜荊荊恩聯(lián)考,15)已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=2,an+1=Sn+2.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=log2an2-11,求數(shù)列{|bn|}的前n項(xiàng)和T解析(1)由an+1=Sn+2,得當(dāng)n≥2時(shí)an=Sn-1+2,兩式相減得an+1-an=an,所以an+1=2an(n≥2).將a1=2代入an+1=Sn+2得,a2=4=2a1,所以對(duì)任意n∈N*,an+1=2an,故{an}是首項(xiàng)為2,公比為2的等比數(shù)列,所以an=2n.(2)bn=log2an2-11=2n令數(shù)列{bn}的前n項(xiàng)和為Bn.Bn=b1+b2+…+bn=n(?9+2n?11)2=n(n-10)=n因?yàn)楫?dāng)n≤5時(shí),bn<0,當(dāng)n≥6時(shí),bn>0,所以當(dāng)n≤5時(shí),Tn=-b1-b2-…-bn=-Bn=10n-n2,當(dāng)n≥6時(shí),Tn=-b1-b2-…-b5+b6+b7+…+bn=Bn-2B5=n2-10n+50.故Tn=10能力拔高練1.(2025屆江蘇如皋開學(xué)考,5)在等差數(shù)列{an}中,若a4+2a9=a2=8,則下列說法錯(cuò)誤的是()A.a1=9B.S10=45C.Sn的最大值為45D.滿足Sn>0的n的最大值為19答案D2.(2025屆廣東廣州三校期中聯(lián)考,7)在等差數(shù)列{an}中,Sn是{an}的前n項(xiàng)和,若S18<0,S19>0,則有限項(xiàng)數(shù)列S1a1,S2a2,…,S18a18A.S9a9,S18a18答案B3.(2025屆湖南永州一模,7)已知數(shù)列{an}滿足an+1?anan=an+2?an+1an+2(n∈N*),且a1=1,a2024=22025,則aA.n答案D4.(2025屆廣東五校開學(xué)聯(lián)考,7)若f(x)=(x-1)3+2(x-1)-lnx2?x+2,數(shù)列{an}的前n項(xiàng)和為Sn,且S1=110,2Sn=nan+1,則i=119[f(ai)+f(a20-i)]A.76B.38C.19D.0答案A5.(多選)(2025屆江蘇南京六校學(xué)情調(diào)研,10)已知等差數(shù)列{an}的首項(xiàng)為a1,公差為d,其前n項(xiàng)和為Sn,若S8<S6<S7,則下列說法正確的是()A.當(dāng)n=7時(shí),Sn最大B.使得Sn<0成立的最小自然數(shù)n=13C.|a6+a7|<|a8+a9|D.數(shù)列S答案ACD6.(多選)(2025屆浙江名校聯(lián)盟開學(xué)考,9)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且公差d≠0,2a15+a18=24.則以下結(jié)論正確的是()A.a16=8B.若S9=S10,則d=4C.若d=-2,則Sn的最大值為S21D.若a15,a16,a18成等比數(shù)列,則d=4答案ABD7.(2025屆江蘇泰州多校聯(lián)考,18)已知數(shù)列{an}為等差數(shù)列,公差d≠0,前n項(xiàng)和為Sn,a2為a1和a5的等比中項(xiàng),S11=121.(1)求數(shù)列{an}的通項(xiàng)公式;(2)是否存在正整數(shù)m,n(3<m<n),使得1a3,1am,1an成等差數(shù)列?若存在,求出m,n(3)求證:i=2解析(1)由題意得a22=a1a5,即(a1+d)2=a1(a1+4整理得d2=2a1d,因?yàn)閐≠0,所以d=2a1,由S11=11a1+55d=121,即11a1+55×2a1=121,解得a1=1,故d=2a1=2,故{an}的通項(xiàng)公式為an=1+2(n-1)=2n-1.(2)不存在正整數(shù)m,n(3<m<n),使得1a3,1am理由如下:假設(shè)存在正整數(shù)m,n(3<m<n),使得1a3,1am因?yàn)?a3=15所以由題意得22m?1=15+12n?1,即5(2n故2m-1與n+2中至少有一個(gè)為5的倍數(shù),不妨設(shè)2m-1為5的倍數(shù),設(shè)2m-1=5t,t∈N*,因?yàn)閙>3,故m的最小值為8,此時(shí)t=3,故t≥3,5(2n-1)=5t(n+2),故n=?1?2tt?2<0若n+2為5的倍數(shù),設(shè)n+2=5k,k∈N*,因?yàn)閚>m>3,所以n的最