新教材2023-2024學(xué)年高中數(shù)學(xué)第五章數(shù)列5.2等差數(shù)列5.2.1等差數(shù)列分層作業(yè)課件新人教B版選擇性必修第三冊(cè)

第五章5.2.1等差數(shù)列A級(jí)必備知識(shí)基礎(chǔ)練12345678910111213141516171.[探究點(diǎn)二]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)數(shù)A解析

令an=90-2n>0,解得n<45.又因?yàn)閚∈N+,所以數(shù)列{an}中正數(shù)項(xiàng)的個(gè)數(shù)為44.12345678910111213141516172.[探究點(diǎn)三]在等差數(shù)列{an}中,已知a4+a8=16,則a2+a10=(

)A.12 B.16

C.20

D.24B解析

a2+a10=a4+a8=16.故選B.12345678910111213141516173.[探究點(diǎn)三]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.860B解析

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a37+b37=a1+b1+36(d1+d2)=100+20×36=820.故選B.12345678910111213141516174.[探究點(diǎn)三]若等差數(shù)列的前3項(xiàng)依次是,則該數(shù)列的公差d是

.

12345678910111213141516175.[探究點(diǎn)三]等差數(shù)列{an}中,若a2,a2022為方程x2-10x+16=0的兩根,則a1+a1012+a2023=

.

15解析

∵a2,a2

022為方程x2-10x+16=0的兩根,∴a2+a2

022=10,∴2a1

012=10,即a1

012=5,∴a1+a1

012+a2

023=3a1

012=15.12345678910111213141516176.[探究點(diǎn)二·2023江蘇泰州高三期末]寫(xiě)出一個(gè)公差不為零,且滿(mǎn)足a1+a2-a3=1的等差數(shù)列{an}的通項(xiàng)公式an=

.

n+1解析

設(shè)等差數(shù)列{an}的公差為d,則a1+a2-a3=a1+a1+d-a1-2d=a1-d=1,不妨令d=1,則a1=2,此時(shí)等差數(shù)列{an}的通項(xiàng)公式為an=n+1.12345678910111213141516177.[探究點(diǎn)二、三](1)求等差數(shù)列8,5,2,…的第20項(xiàng);(2)[人教A版教材習(xí)題]已知{an}為等差數(shù)列,a1+a3+a5=105,a2+a4+a6=99,求a20.解

(1)設(shè)等差數(shù)列為數(shù)列{an}且其公差為d,則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{an}的公差為d.由題意,知a1+a3+a5=3a3=105,所以a3=35.由a2+a4+a6=3a4=99,得a4=33,所以d=a4-a3=-2,所以a20=a3+17d=35+17×(-2)=1.12345678910111213141516178.[探究點(diǎn)一]已知數(shù)列{an}的各項(xiàng)都為正數(shù),前n項(xiàng)和為Sn,且

.(1)求a1,a2;(2)求證:數(shù)列{an}是等差數(shù)列.解得a2=-1(舍去)或a2=3.1234567891011121314151617∴(an+an-1)(an-an-1-2)=0.∵an>0,∴an+an-1≠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.1234567891011121314151617B級(jí)關(guān)鍵能力提升練9.在數(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(

)B解析

令bn=an+n2.∵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-1,1234567891011121314151617又n∈N+,故選B.123456789101112131415161710.(多選題)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,則下列能判斷數(shù)列{an}是等差數(shù)列的是(

)A.Sn=nB.Sn=n2+nC.Sn=2nD.Sn=n2+n+1AB1234567891011121314151617解析

對(duì)于A,當(dāng)n≥2時(shí),an=Sn-Sn-1=n-(n-1)=1,而a1=S1=1滿(mǎn)足上式,則an=1(n∈N+),數(shù)列{an}是常數(shù)數(shù)列,是等差數(shù)列;對(duì)于B,當(dāng)n≥2時(shí),an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n,而a1=S1=2滿(mǎn)足上式,則an=2n(n∈N+),數(shù)列{an}是等差數(shù)列;對(duì)于C,當(dāng)n≥2時(shí),an=Sn-Sn-1=2n-2n-1=2n-1,而a1=S1=2不滿(mǎn)足上式,對(duì)于D,當(dāng)n≥2時(shí),an=Sn-Sn-1=n2+n+1-(n-1)2-(n-1)-1=2n,而a1=S1=3不滿(mǎn)足上式,123456789101112131415161711.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

.

an=38-5n解析

設(shè)等差數(shù)列{an}的公差為d.又d∈Z,∴d=-5,∴an=33+(n-1)×(-5)=38-5n.123456789101112131415161712.[2023山東煙臺(tái)高二專(zhuān)題練習(xí)]已知數(shù)列{an}是等差數(shù)列,數(shù)列{bn}滿(mǎn)足bn=an-2n,b3=-1,b5=-21,則{an}的公差d為

.

2解析

由bn=an-2n得an=bn+2n,則a3=b3+8=-1+8=7,a5=b5+32=-21+32=11,則123456789101112131415161713.同時(shí)滿(mǎn)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.n-2解析

設(shè)等差數(shù)列{an}的公差為d,由①可知d>0.由a2-a3+a4=1,得a3=a1+2d=1.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-2.123456789101112131415161714.四個(gè)數(shù)成遞增的等差數(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).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-2,0,2,4.1234567891011121314151617(2)求數(shù)列{an}的通項(xiàng)公式.123456789101112131415161716.已知數(shù)列{an}滿(mǎn)足a1=2,an=2an-1+2n+1(n≥2,n∈N+).設(shè)

求證:數(shù)列{bn}是等差數(shù)列,并分別求an和bn.∴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,∴an=(2n-1)·2n.123456789101112131415161717.已知等差數(shù)列{an}:3,7,11,15,….(1)求{an}的通項(xiàng)公式.(2)135,4m+19(m∈N+)是數(shù)列{an}中的項(xiàng)嗎?如果是,是第幾項(xiàng)?(3)若as,at(s,t∈N+)是數(shù)列{an}中的項(xiàng),那么2as+3at是數(shù)列{an}中的項(xiàng)嗎?如果是,是第幾項(xiàng)?C級(jí)學(xué)科素養(yǎng)創(chuàng)新練解

(1)設(shè)數(shù)列{an}的公差為d.依題意,有a1=3,d=7-3=4,∴an=3+4(n-1)=4n-1.(2)由(1)可知,{an}的通項(xiàng)公式為an=4n-1.令4n-1=135,得n=34,∴135是數(shù)列{an}的第34項(xiàng).∵4m+19=4(m+5)-1,且m∈N+,∴4m+19(m∈N+)是數(shù)列{an}的第(m+5)項(xiàng).12345678910111213