歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！5.2等差數(shù)列5.2.1等差數(shù)列必備知識(shí)基礎(chǔ)練1.已知等差數(shù)列{an}的通項(xiàng)公式為an=90-2n,則這個(gè)數(shù)列中正數(shù)項(xiàng)的個(gè)數(shù)為()A.44 B.45 C.90 D.無(wú)窮多2.在等差數(shù)列{an}中,已知a4+a8=16,則a2+a10=()A.12 B.16 C.20 D.243.(2022江蘇鹽城三模)已知數(shù)列{an},{bn}均為等差數(shù)列,且a1=25,b1=75,a2+b2=120,則a37+b37的值為()A.760 B.820 C.780 D.8604.若等差數(shù)列的前3項(xiàng)依次是1x+1,565.等差數(shù)列{an}中,若a2,a2022為方程x2-10x+16=0的兩根,則a1+a1012+a2023=.

6.一種游戲軟件的租金,第一天6元,第二天12元,以后每天比前一天多3元,則第n(n≥2)天的租金an=(單位:元).

7.(1)求等差數(shù)列8,5,2,…的第20項(xiàng);(2)-401是不是等差數(shù)列-5,-9,-13,…中的項(xiàng)?如果是,那么是第幾項(xiàng)?8.已知數(shù)列{an}的各項(xiàng)都為正數(shù),前n項(xiàng)和為Sn,且Sn=14(an+1)2(n∈N+)(1)求a1,a2;(2)求證:數(shù)列{an}是等差數(shù)列.關(guān)鍵能力提升練9.數(shù)列{an}中,a1=5,a2=9.若數(shù)列{an+n2}是等差數(shù)列,則數(shù)列{an}的最大項(xiàng)為()A.9 B.11 C.454 10.設(shè)數(shù)列{an}是首項(xiàng)為50,公差為2的等差數(shù)列,數(shù)列{bn}是首項(xiàng)為10,公差為4的等差數(shù)列,以ak和bk為兩邊的矩形內(nèi)的最大圓的面積記為Sk,如果k≤21,那么Sk等于()A.π(k+24)2 B.π(k+12)2C.π(2k+3)2 D.π(2k+1)211.(2022北京鐵路二中高二期中)在等差數(shù)列{an}中,首項(xiàng)為33,公差為整數(shù),若前7項(xiàng)均為正數(shù),第7項(xiàng)以后各項(xiàng)均為負(fù)數(shù),則數(shù)列的通項(xiàng)公式為.

12.已知數(shù)列{an}的首項(xiàng)a1=21,且滿足(2n-5)an+1=(2n-3)an+4n2-16n+15,則數(shù)列{an}的最小項(xiàng)是第項(xiàng).

13.(2022河南商丘三模)同時(shí)滿足下面兩個(gè)性質(zhì)的數(shù)列{an}的一個(gè)通項(xiàng)公式為an=.

①是遞增的等差數(shù)列;②a2-a3+a4=1.14.四個(gè)數(shù)成遞增等差數(shù)列,中間兩數(shù)的和為2,首末兩項(xiàng)的積為-8,求這四個(gè)數(shù).15.(2022遼寧撫順高級(jí)中學(xué)高二階段練習(xí))已知各項(xiàng)均為正數(shù)的數(shù)列{an}中,a1=12,an-an+1=2anan+1(1)證明:數(shù)列1a(2)求數(shù)列{an}的通項(xiàng)公式.16.已知數(shù)列{an}滿足a1=2,an=2an-1+2n+1(n≥2,n∈N+).(1)設(shè)bn=an2n,求證數(shù)列{(2)若數(shù)列{cn}滿足cn=2n+1,且對(duì)于任意正整數(shù)n,不等式abn+4≤1+1c11+1c2…1+1c學(xué)科素養(yǎng)創(chuàng)新練17.數(shù)列{an}滿足a1=1,an+1=(n2+n-λ)an(n=1,2,…),λ是常數(shù).(1)當(dāng)a2=-1時(shí),求λ及a3的值.(2)是否存在實(shí)數(shù)λ使數(shù)列{an}為等差數(shù)列?若存在,求出λ及數(shù)列{an}的通項(xiàng)公式;若不存在,請(qǐng)說(shuō)明理由.參考答案5.2等差數(shù)列5.2.1等差數(shù)列1.A令an=90-2n>0,解得n<45.又因?yàn)閚∈N+,所以n=44.故數(shù)列{an}中正數(shù)項(xiàng)的個(gè)數(shù)為44.2.Ba2+a10=a4+a8=16,故選B.3.B設(shè)等差數(shù)列{an},{bn}的公差分別為d1,d2.因?yàn)閍1+b1=100,a2+b2=120,所以d1+d2=120-100=20,所以數(shù)列{an+bn}也為等差數(shù)列,且首項(xiàng)為100,公差為20,所以a37+b37=100+20×36=820.故選B.4.112依題意得2×56x=1x+1+5.15∵a2,a2022為方程x2-10x+16=0的兩根,∴a2+a2022=10,∴2a1012=10,即a1012=5,∴a1+a1012+a2023=3a1012=15.6.3n+6(n≥2)a1=6,a2=12,a3=15,a4=18,……,從第2項(xiàng)起,數(shù)列{an}中的項(xiàng)才構(gòu)成等差數(shù)列,且公差為3,在這個(gè)等差數(shù)列中第1項(xiàng)是12,而第n天的租金,是第(n-1)項(xiàng),故an=12+(n-2)×3=3n+6(n≥2).7.解(1)設(shè)等差數(shù)列為數(shù)列{an}且其公差為d,則a1=8,d=5-8=-3,得數(shù)列的通項(xiàng)公式為an=-3n+11,所以a20=-49.(2)設(shè)等差數(shù)列為數(shù)列{bn}且其公差為d,則b1=-5,d=-9-(-5)=-4,得數(shù)列通項(xiàng)公式為bn=-5-4(n-1)=-4n-1.令bn=-401,解得n=100,即-401是這個(gè)數(shù)列的第100項(xiàng).8.(1)解由已知條件得,a1=14(a1+1)2∴a1=1.又有a1+a2=14(a2+1)2,即a22-2a2-3解得a2=-1(舍)或a2=3.(2)證明由Sn=14(an+1)2當(dāng)n≥2時(shí),Sn-1=14(an-1+1)2∴Sn-Sn-1=14[(an+1)2-(an-1+1)2]=14[an2?a即4an=an2?an-12+2∴an2?an-12-2a∴(an+an-1)(an-an-1-2)=0,∴an-an-1-2=0,即an-an-1=2(n≥2),∴數(shù)列{an}是首項(xiàng)為1,公差為2的等差數(shù)列.9.B令bn=an+n2,又a1=5,a2=9,∴b1=a1+1=6,b2=a2+4=13,∴數(shù)列{an+n2}的公差為13-6=7,則an+n2=6+7(n-1)=7n-1,∴an=-n2+7n-1=-n-722+454.又n∈N+,∴當(dāng)n=3或n=4時(shí),an有最大值為-14+故選B.10.C由題意,得ak=2k+48,bk=4k+6,bk-ak=(4k+6)-(2k+48)=2k-42.∵k≤21,∴2k-42≤0,∴bk≤ak,∴矩形內(nèi)的最大圓是以bk為直徑的圓.因此Sk=π(2k+3)2.11.an=38-5n由題意可得a即33+6解得-336<d<-33又d∈Z,∴d=-5,∴an=33+(n-1)×(-5)=38-5n.12.5易知(2n-3)(2n-5)≠0,故由已知得an+12n-所以數(shù)列an2n-5是首項(xiàng)為-7,公差為所以an2n-5=-7+則an=(2n-5)(n-8),函數(shù)y=(2x-5)(x-8)的圖象的對(duì)稱軸為直線x=10.52=5.25,所以數(shù)列{an}的最小項(xiàng)是第13.n-2(答案不唯一,滿足d>0,a3=1即可)設(shè)等差數(shù)列{an}的公差為d.由a2-a3+a4=1,得a3=a1+2d=1.由①可知d>0,取d=1,則a1=-1,所以數(shù)列{an}的一個(gè)通項(xiàng)公式為an=-1+(n-1)=n-2.14.解設(shè)這四個(gè)數(shù)為a-3d,a-d,a+d,a+3d(公差為2d),依題意,2a=2,且(a-3d)(a+3d)=-8,即a=1,a2-9d2=-8,∴d2=1,∴d=1或d=-1.又四個(gè)數(shù)成遞增的等差數(shù)列,∴d>0,∴d=1,故所求的四個(gè)數(shù)為-2,0,2,4.15.(1)證明由已知得1a1=2,an≠0,1所以數(shù)列1an是以2為首項(xiàng),2(2)解由(1)知,1an=1a1+2(所以an=1216.(1)證明∵an=2an-1+2n+1,∴an2n=an-12n∵bn=an∴bn=bn-1+2(n≥2,n∈N+).又b1=a12∴數(shù)列{bn}是以1為首項(xiàng),2為公差的等差數(shù)列,∴bn=1+(n-1)×2=2n-1.(2)解由abn+4≤1+1c11+1c2…1+1c∵cn=2n+1,∴1+1cn>記f(n)=12n+31+1c11+1c2…則f(n+1又f(n)>0,∴f(n+1)>f(n),即f(n)在N+上單調(diào)遞增.故f(n)min=f(1)=45∴0<a≤45即a的取值范圍