1、第1講等差數(shù)列、等比數(shù)列A組基礎(chǔ)題組1.(2017湖南五市十校聯(lián)考)已知Sn是數(shù)列an的前n項(xiàng)和,且Sn+1=Sn+an+3,a4+a5=23,則S8=() A.72B.88C.92D.982.(2017西安八校聯(lián)考)已知數(shù)列an是等比數(shù)列,數(shù)列bn是等差數(shù)列,若a1a6a11=-3,b1+b6+b11=7,則tan的值是()A.-B.-1C.-D.3.(2017張掖第一次診斷考試)等差數(shù)列an中,是一個(gè)與n無關(guān)的常數(shù),則該常數(shù)的可能值的集合為()A.1B.C.D.4.在等比數(shù)列an中,公比q=2,前87項(xiàng)和S87=140,則a3+a6+a9+a87等于()A.B.60C.80D.1605.已

2、知數(shù)列an,則有()A.若=4n,nN*,則an為等比數(shù)列B.若anan+2=,nN*,則an為等比數(shù)列C.若aman=2m+n,m,nN*,則an為等比數(shù)列D.若an=an+1an+2,nN*,則an為等比數(shù)列6.(2017鄭州第二次質(zhì)量預(yù)測)在等差數(shù)列an中,an0,a7=a4+4,Sn為數(shù)列an的前n項(xiàng)和,S19=.7.設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,S3=6,S4=12,則S6=.8.設(shè)某數(shù)列的前n項(xiàng)和為Sn,若為常數(shù),則稱該數(shù)列為“和諧數(shù)列”.若一個(gè)首項(xiàng)為1,公差為d(d0)的等差數(shù)列an為“和諧數(shù)列”,則該等差數(shù)列的公差d=.9.已知等比數(shù)列an的前n項(xiàng)和為Sn,且滿足S3=,S6

3、=.(1)求數(shù)列an的通項(xiàng)公式;(2)求log2a1+log2a2+log2a3+log2a25的值.10.(2017云南第一次統(tǒng)一檢測)已知數(shù)列an中,+2an-n2+2n=0.(1)求數(shù)列an的通項(xiàng)公式;(2)求數(shù)列an的前n項(xiàng)和Sn.B組提升題組1.(2017鄭州第一次質(zhì)量預(yù)測)已知數(shù)列an滿足a1a2a3an=(nN*),且對任意nN*都有+t,則實(shí)數(shù)t的取值范圍為()A.B.C.D.2.(2017福建普通高中質(zhì)量檢查)已知數(shù)列an滿足a1=-40,且nan+1-(n+1)an=2n2+2n,則an取最小值時(shí)n的值為.3.已知數(shù)列an滿足:an+1-an=d(nN*),前n項(xiàng)和記為Sn

4、,a1=4,S3=21.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列bn滿足b1=,bn+1-bn=,求數(shù)列bn的通項(xiàng)公式.4.已知數(shù)列an是等差數(shù)列,滿足a2=5,a4=13,數(shù)列bn的前n項(xiàng)和是Tn,且Tn+bn=3.(1)求數(shù)列an及數(shù)列bn的通項(xiàng)公式;(2)設(shè)cn=anbn,求數(shù)列cn中的最大項(xiàng).答案精解精析A組基礎(chǔ)題組1.C解法一:由Sn+1=Sn+an+3得an+1-an=3,則數(shù)列an是公差為3的等差數(shù)列,又a4+a5=23=2a1+7d=2a1+21,a1=1,S8=8a1+d=92.解法二:由Sn+1=Sn+an+3得an+1-an=3,則數(shù)列an是公差為3的等差數(shù)列,S8=92

5、.2.A依題意得,=(-)3,3b6=7,a6=-,b6=,=-,故tan=tan=tan=-tan=-,選A.3.B=,若a1=d,則=;若a10,d=0,則=1.a1=d0,0,該常數(shù)的可能值的集合為.4.C解法一:a3+a6+a9+a87=a3(1+q3+q6+q84)=a1q2=140=80.故選C.解法二:設(shè)b1=a1+a4+a7+a85,b2=a2+a5+a8+a86,b3=a3+a6+a9+a87,因?yàn)閎1q=b2,b2q=b3,且b1+b2+b3=140,所以b1(1+q+q2)=140,而1+q+q2=7,所以b1=20,b3=q2b1=420=80.故選C.5.C若a1=-

6、2,a2=4,a3=8,滿足=4n,nN*,但an不是等比數(shù)列,故A錯(cuò);若an=0,滿足anan+2=,nN*,但an不是等比數(shù)列,故B錯(cuò);若an=0,滿足anan+3=an+1an+2,nN*,但an不是等比數(shù)列,故D錯(cuò);aman=2m+n,m,nN*,則有=2,則an是等比數(shù)列.6.答案152解析由a7=a4+4得a1+6d=(a1+3d)+4,即a1+9d=8,所以a10=8,因此S19=19a10=198=152.7.答案30解析設(shè)數(shù)列an的公差為d,S3=6,S4=12,S6=6a1+d=30.8.答案2解析由=k(k為常數(shù)),且a1=1,得n+n(n-1)d=k,即2+(n-1)d

7、=4k+2k(2n-1)d,整理得,(4k-1)dn+(2k-1)(2-d)=0,對任意正整數(shù)n,上式恒成立,得數(shù)列an的公差為2.9.解析(1)設(shè)公比為q,由題意可得,公比q1,再由S3=,S6=可得解得故通項(xiàng)公式為an=2n-1=2n-2.(2)由(1)及題意可得log2an=n-2,所以log2a1+log2a2+log2a3+log2a25=-1+0+1+2+23=275.10.解析(1)由+2an-n2+2n=0得(an-n+2)(an+n)=0,an=n-2或an=-n.an的通項(xiàng)公式為an=n-2或an=-n.(2)當(dāng)an=n-2時(shí),易知an為等差數(shù)列,且a1=-1,Sn=.當(dāng)an=-n時(shí),易知an也為等差數(shù)列,且a1=-1,Sn=-.B組提升題組1.D數(shù)列an滿足a