新教材2023-2024學(xué)年高中數(shù)學(xué)第1章數(shù)列1.2等差數(shù)列1.2.1等差數(shù)列及其通項(xiàng)公式第1課時(shí)等差數(shù)列的概念及通項(xiàng)公式分層作業(yè)課件湘教版選擇性必修第一冊

第1章1.2.1第1課時(shí)等差數(shù)列的概念及通項(xiàng)公式A級必備知識基礎(chǔ)練12345678910111213141.在等差數(shù)列{an}中,若a1=2,a2=4,則a4=(

)A.6 B.8

C.16

D.32B解析

因?yàn)樵诘炔顢?shù)列{an}中,a1=2,a2=4,所以公差d=a2-a1=4-2=2,則a4=a1+3d=2+3×2=8,故選B.12345678910111213142.若數(shù)列{an}滿足a1=19,an+1=an-3(n∈N+),則當(dāng)am=-2021時(shí),m的值是(

)A.679 B.680

C.681

D.690C解析

∵a1=19,an+1-an=-3(n∈N+),∴{an}是以19為首項(xiàng),-3為公差的等差數(shù)列,則an=19+(n-1)×(-3)=22-3n.令am=22-3m=-2

021,解得m=681.故選C.12345678910111213143.在等差數(shù)列{an}中,a3+a9=32,a2=4,則a10=(

)A.25 B.28

C.31

D.34B解析

因?yàn)樵诘炔顢?shù)列{an}中,a3+a9=32,a2=4,所以2a1+10d=32,a1+d=4,解得a1=1,d=3,所以a10=a1+9d=28,故選B.12345678910111213144.若等差數(shù)列{an}的公差d=2,a8∶a7=7∶8,則a1=(

)A.-15 B.-28 C.15

D.28B解析

設(shè)a8=7k,a7=8k,k∈R,則a8-a7=7k-8k=-k=2,則k=-2.即a7=-16,故a1=a7-6d=-16-12=-28,故選B.1234567891011121314B12345678910111213146.(多選題)已知數(shù)列{an}為等差數(shù)列,則下列說法正確的是(

)A.數(shù)列{an+b}(b為常數(shù))是等差數(shù)列B.數(shù)列{-an}是等差數(shù)列C.數(shù)列{}是等差數(shù)列D.an+1是an與an+2的等差中項(xiàng)ABD1234567891011121314解析

記數(shù)列{an}的公差為d,則an+1-an=d.(an+1+b)-(an+b)=d,故A正確;(-an+1)-(-an)=-(an+1-an)=-d,所以數(shù)列{-an}是等差數(shù)列,故B正確;根據(jù)等差數(shù)列的定義可知2an+1=an+an+2,所以an+1是an與an+2的等差中項(xiàng),故D正確.故選ABD.12345678910111213147.已知等差數(shù)列{an}的前3項(xiàng)依次是-1,a-1,1,則a=

,通項(xiàng)公式an=

.

1n-2解析

因?yàn)?1,a-1,1構(gòu)成等差數(shù)列,所以2(a-1)=-1+1=0,解得a=1.因?yàn)閍1=-1,d=1,所以an=n-2.12345678910111213148.若a≠b,則等差數(shù)列a,x1,x2,b的公差是(

)B級關(guān)鍵能力提升練C解析

由等差數(shù)列的通項(xiàng)公式,得b=a+(4-1)d,所以d=.12345678910111213149.在數(shù)列{an}中,a1=-1,a3=3,an+2=2an+1-an(n∈N+),則a10=(

)A.10 B.17

C.21

D.35B解析

∵an+2=2an+1-an(n∈N+),∴an+2+an=2an+1,即數(shù)列{an}是等差數(shù)列.設(shè){an}的公差為d,∵a1=-1,a3=3,∴a3=a1+2d,即3=-1+2d,得d=2.則an=-1+(n-1)×2=2n-3,故a10=2×10-3=17.1234567891011121314A.121 B.144

C.169

D.196C1234567891011121314123456789101112131412.已知數(shù)列{an}滿足a1=15,且3an+1=3an-2(n∈N+).若akak+1<0,則數(shù)列的通項(xiàng)公式為

,正整數(shù)k=

.

23123456789101112131413.已知數(shù)列{an}滿足an=2an-1+2n-1(n≥2),且a4=81.(1)求數(shù)列{an}的前三項(xiàng)a1,a2,a3.(2)是否存在一個(gè)實(shí)數(shù)λ,使得數(shù)列{}為等差數(shù)列?若存在,求出λ的值;若不存在,請說明理由.(3)求數(shù)列{an}的通項(xiàng)公式.解

(1)由題意知a4=2a3+24-1=81,解得a3=33.同理可得a2=13,a1=5.12345678910111213141234567891011121314C級學(xué)科素養(yǎng)創(chuàng)新練14.(多選題)已知等差數(shù)列{an}是無窮數(shù)列,該數(shù)列的首項(xiàng)a1=3,公差d=-5,依次取出項(xiàng)的序號被4除余3的項(xiàng)組成數(shù)列{bn},則(

)A.b1=-7B.b2=27C.an=8-5nD.{bn}中的第503項(xiàng)是{an}中的第2020項(xiàng)AC

1234567891011121314解析

∵a1=3,d=-5,∴an=3+(n-1)×(-5)=8-5n,故C正確;數(shù)列{an}中項(xiàng)的序號被4除余3的項(xiàng)是第3項(xiàng),第7項(xiàng),第11項(xiàng),…,∴b1=a3=-7,b2=a7=-27,故A正確,B錯(cuò)