2025屆高考數(shù)學(xué)一輪復(fù)習(xí)第5章數(shù)列第2講等差數(shù)列及其前n項(xiàng)和作業(yè)試題1含解析新人教版

PAGE第五章數(shù)列其次講等差數(shù)列及其前n項(xiàng)和練好題﹒考點(diǎn)自測(cè)1.[2024全國(guó)Ⅰ,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.-12 B.-10 C.10 D.122.[2024全國(guó)卷Ⅱ,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(chēng)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)比上一層的最終一環(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3.[2024浙江,4分]已知等差數(shù)列{an}的前n項(xiàng)和為Sn,公差d≠0,且QUOTE≤1.記b1=S2,bn+1=S2n+2-S2n,n∈N*,下列等式不行能成立的是 ()A.2a4=a2+a6 B.2b4=b2+b6C.QUOTE=a2a8 D.QUOTE=b2b84.[2024北京,4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)5.[多選題]下面結(jié)論正確的是 ()A.若一個(gè)數(shù)列從第2項(xiàng)起每一項(xiàng)與它的前一項(xiàng)的差都是常數(shù),則這個(gè)數(shù)列是等差數(shù)列B.數(shù)列{an}為等差數(shù)列的充要條件是對(duì)隨意n∈N*,都有2an+1=an+an+2C.數(shù)列{an}為等差數(shù)列的充要條件是其通項(xiàng)公式為n的一次函數(shù)D.已知數(shù)列{an}的通項(xiàng)公式是an=pn+q(其中p,q為常數(shù)),則數(shù)列{an}肯定是等差數(shù)列6.[2024山東,5分]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.

7.[2024北京,5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.

8.一個(gè)等差數(shù)列的前12項(xiàng)和為354,前12項(xiàng)中偶數(shù)項(xiàng)的和與奇數(shù)項(xiàng)的和的比為32∶27,則該數(shù)列的公差d=.

拓展變式1.[2024石家莊二檢]已知數(shù)列{an}中,a1=1,當(dāng)n≥2時(shí),an-1-an=an-1·an.(1)求證:數(shù)列{QUOTE}是等差數(shù)列.(2)設(shè)bn=a2n-1·a2n+1,數(shù)列{bn}的前n項(xiàng)和為T(mén)n,求證:Tn<QUOTE.2.[2024青島市5月模擬][多選題]已知等差數(shù)列{an}的前n項(xiàng)和為Sn(n∈N*),公差d≠0,S6=90,a7是a3與a9的等比中項(xiàng),則下列選項(xiàng)正確的是 ()A.a1=22 B.d=-2C.當(dāng)n=10或n=11時(shí),Sn取得最大值 D.當(dāng)Sn>0時(shí),n的最大值為203.(1)[2024貴陽(yáng)市摸底測(cè)試]等差數(shù)列{an}的前n項(xiàng)和為Sn,若S17=51,則2a10-a11= ()A.2 B.3 C.4 D.6(2)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若m>1,且am-1+am+1-QUOTE-1=0,S2m-1=39,則m等于()A.39 B.20 C.19 D.10(3)等差數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,若QUOTE=QUOTE,則QUOTE=.

4.[2024全國(guó)卷Ⅱ,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.答案其次講等差數(shù)列及其前n項(xiàng)和1.B解法一設(shè)等差數(shù)列{an}的公差為d,∵3S3=S2+S4,∴3(3a1+QUOTEd)=2a1+d+4a1+QUOTEd,解得d=-QUOTEa1.∵a1=2,∴d=-3,∴a5=a1+4d=2+4×(-3)=-10.故選B.解法二設(shè)等差數(shù)列{an}的公差為d,∵3S3=S2+S4,∴3S3=S3-a3+S3+a4,∴S3=a4-a3,∴3a1+QUOTEd=d,又a1=2,∴d=-3,∴a5=a1+4d=2+4×(-3)=-10.故選B.2.C由題意知,由天心石起先向外的每環(huán)的扇面形石板塊數(shù)構(gòu)成一個(gè)等差數(shù)列,記為{an},設(shè)數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,易知其首項(xiàng)a1=9,d=9,所以an=a1+(n-1)d=9n.由等差數(shù)列的性質(zhì)知Sn,S2n-Sn,S3n-S2n也成等差數(shù)列,所以2(S2n-Sn)=Sn+S3n-S2n,所以(S3n-S2n)-(S2n-Sn)=S2n-2Sn=QUOTE-2×QUOTE=9n2=729,得n=9,所以三層共有扇面形石板的塊數(shù)為S3n=QUOTE=QUOTE=3402,故選C.3.D由bn+1=S2n+2-S2n,得b2=a3+a4=2a1+5d,b4=a7+a8=2a1+13d,b6=a11+a12,b8=a15+a16=2a1+29d.由等差數(shù)列的性質(zhì)易知A成立;若2b4=b2+b6,則2(a7+a8)=a3+a4+a11+a12=2a7+2a8,故B成立;若QUOTE=a2a8,即(a1+3d)2=(a1+d)(a1+7d),則a1=d,故C可能成立;若QUOTE=b2b8,即QUOTE=(2a1+5d)(2a1+29d),則QUOTE=QUOTE,與已知沖突,故D不行能成立.4.B設(shè)等差數(shù)列{an}的公差為d,∵a1=-9,a5=-1,∴a5=-9+4d=-1,∴d=2,∴an=-9+(n-1)×2=2n-11.令an=2n-11≤0,則n≤5.5.∴n≤5時(shí),an<0;n≥6時(shí),an>0.∴T1=-9<0,T2=(-9)×(-7)=63>0,T3=(-9)×(-7)×(-5)=-315<0,T4=(-9)×(-7)×(-5)×(-3)=945>0,T5=(-9)×(-7)×(-5)×(-3)×(-1)=-945<0,當(dāng)n≥6時(shí),an>0,且an≥1,∴Tn+1<Tn<0,∴Tn=a1a2…an(n=1,2,…)有最大項(xiàng)T4,無(wú)最小項(xiàng),故選B.5.BD對(duì)于A,若一個(gè)數(shù)列從第2項(xiàng)起每一項(xiàng)與它的前一項(xiàng)的差都是同一個(gè)常數(shù),則這個(gè)數(shù)列是等差數(shù)列,故A錯(cuò)誤;對(duì)于B,由2an+1=an+an+2得an+1-an=an+2-an+1,故B正確;對(duì)于C,數(shù)列{an}為等差數(shù)列的充分不必要條件是其通項(xiàng)公式為n的一次函數(shù),故C錯(cuò)誤;對(duì)于D,由等差數(shù)列與一次函數(shù)的關(guān)系可得D正確.故選BD.6.3n2-2n設(shè)bn=2n-1,cn=3n-2,bn=cm,則2n-1=3m-2,得n=QUOTE=QUOTE=QUOTE+1,于是m-1=2k,k∈N,所以m=2k+1,k∈N,則ak=3(2k+1)-2=6k+1,k∈N,得an=6n-5,n∈N*.故Sn=QUOTE×n=3n2-2n.7.0-10設(shè)等差數(shù)列{an}的公差為d,∵QUOTE即QUOTE可得QUOTE∴a5=a1+4d=0.∵Sn=na1+QUOTEd=QUOTE(n2-9n),∴當(dāng)n=4或n=5時(shí),Sn取得最小值,最小值為-10.8.5設(shè)等差數(shù)列的前12項(xiàng)中奇數(shù)項(xiàng)的和為S奇,偶數(shù)項(xiàng)的和為S偶,等差數(shù)列的公差為d.由已知條件,得QUOTE解得QUOTE又S偶-S奇=6d,所以d=QUOTE=5.1.(1)當(dāng)n≥2時(shí),an-1-an=an-1·an,兩邊同時(shí)除以an-1·an,得QUOTE-QUOTE=1,由a1=1,得QUOTE=1,故數(shù)列{QUOTE}是以1為首項(xiàng),1為公差的等差數(shù)列.(2)由(1)知an=QUOTE,所以bn=QUOTE·QUOTE=QUOTE=QUOTE(QUOTE-QUOTE),所以Tn=QUOTE[(1-QUOTE)+(QUOTE-QUOTE)+…+(QUOTE-QUOTE)]=QUOTE(1-QUOTE).因?yàn)镼UOTE>0,所以Tn<QUOTE.2.BCD因?yàn)閧an}是等差數(shù)列,S6=90,a7是a3與a9的等比中項(xiàng),所以QUOTE解得QUOTE故A錯(cuò)誤,B正確,數(shù)列{an}的前n項(xiàng)和Sn=20n+QUOTE×(-2)=-n2+21n=-(n-QUOTE)2+QUOTE,所以當(dāng)n=10或n=11時(shí)Sn取得最大值110,C正確,當(dāng)Sn>0,即-n2+21n>0時(shí),解得0<n<21,又n∈N*,故n的最大值為20,D正確.故選BCD.3.(1)B解法一∵S17=51,∴QUOTE=51,可得a1+a17=6=2a9,解得a9=3,∴2a10-a11=a9+a11-a11=a9=3.故選B.解法二由S17=17a9=51,得a9=3,則2a10-a11=a9+a11-a11=a9=3.故選B.(2)B數(shù)列{an}為等差數(shù)列,則am-1+am+1=2am,則am-1+am+1-QUOTE-1=0可化為2am-QUOTE-1=0,解得am