（新課標(biāo)）2020版高考數(shù)學(xué)總復(fù)習(xí) 第六章 第三節(jié) 等比數(shù)列及其前n項(xiàng)和練習(xí) 文 新人教A版.docx

第三節(jié)等比數(shù)列及其前n項(xiàng)和A組基礎(chǔ)題組1.已知等比數(shù)列an的前n項(xiàng)和為Sn,且a1+a3=52,a2+a4=54,則Snan=() A.4n-1B.4n-1C.2n-1D.2n-1答案D設(shè)等比數(shù)列an的公比為q,a1+a3=52,a2+a4=54,a1+a1q2=52,a1q+a1q3=54,由可得1+q2q+q3=2,q=12,代入解得a1=2,an=212n-1=42n,Sn=21-12n1-12=41-12n,Snan=41-12n42n=2n-1,選D.2.(2019湖北武漢調(diào)研)設(shè)公比為q(q0)的等比數(shù)列an的前n項(xiàng)和為Sn.若S2=3a2+2,S4=3a4+2,則a1=()A.-2B.-1C.12D.23答案B由S2=3a2+2,S4=3a4+2作差可得a3+a4=3a4-3a2,即2a4-a3-3a2=0,所以2q2-q-3=0,解得q=32或q=-1(舍).將q=32代入S2=3a2+2得,a1+32a1=332a1+2,解得a1=-1.3.(2018湖北武漢聯(lián)考)已知an為等比數(shù)列,a4+a7=2,a5a6=-8,則a1+a10=()A.7B.5C.-5D.-7答案D由a4+a7=2,a5a6=-8得a4+a7=2,a4a7=-8,解得a4=4,a7=-2或a4=-2,a7=4.若a4=4,a7=-2,則a1=-8,a10=1,此時(shí)a1+a10=-7;若a4=-2,a7=4,則a1=1,a10=-8,此時(shí)a1+a10=-7.故選D.4.已知等比數(shù)列an的前n項(xiàng)和為Sn=a2n-1+16,則a的值為()A.13B.12C.-13D.-12答案C由題意得a1=a+16,a2=S2-S1=a,a3=S3-S2=2a,又a22=a1a3,所以a2=a+162a,解得a=-13或a=0(舍去),故選C.5.已知遞增的等比數(shù)列an的公比為q,其前n項(xiàng)和Sn0,則()A.a10,0q1B.a11C.a10,0q0,q1答案ASn0,a1an,且|an|an+1|,則-an-an+10,則q=-an+1-an(0,1),a10,0q1,令bn=an+1(n=1,2,),若數(shù)列bn有連續(xù)四項(xiàng)在集合-53,-23,19,37,82中,則q等于()A.-12B.12C.-32D.32答案Cbn有連續(xù)四項(xiàng)在-53,-23,19,37,82中且bn=an+1,則an有連續(xù)四項(xiàng)在-54,-24,18,36,81中.an是等比數(shù)列,等比數(shù)列中有負(fù)數(shù)項(xiàng),則q1,此種情況舍去),q=-32.故選C.2.在遞增的等比數(shù)列an中,已知a1+an=34,a3an-2=64,且前n項(xiàng)和為Sn=42,則n=.答案3解析設(shè)遞增等比數(shù)列an的公比為q(q0),a3an-2=a1an=64,a1+an=34,解得a1=2,an=32,則Sn=a1-anq1-q=42,解得q=4,則qn-1=4n-1=ana1=16,解得n=3.3.(2018四川成都第一次診斷性檢測(cè))已知數(shù)列an滿足a1=-2,an+1=2an+4.(1)證明數(shù)列an+4是等比數(shù)列;(2)求數(shù)列|an|的前n項(xiàng)和Sn.解析(1)證明:因?yàn)閍1=-2,所以a1+4=2.因?yàn)閍n+1=2an+4,所以an+1+4=2an+8=2(an+4),所以an+1+4an+4=2,所以an+4是以2為首項(xiàng),2為公比的等比數(shù)列.(2)由(1)可知an+4=2n,所以an=2n-4.當(dāng)n2時(shí),an0,所以Sn=-a1+a2+an=2+(22-4)+(2n-4)=2+22+2n-4(n-1)=2(1-2n)1-2-4(n-1)=2n+1-4n+2.又當(dāng)n=1時(shí),上式也滿足.所以當(dāng)nN*時(shí),Sn=2n+1-4n+2.4.(2019云南昆明質(zhì)檢)已知an是等差數(shù)列,滿足a1=3,a4=12,數(shù)列bn滿足b1=4,b4=20,且bn-an為等比數(shù)列.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)求數(shù)列bn的前n項(xiàng)和.解析(1)設(shè)等差數(shù)列an的公差為d,由題意得d=a4-a13=12-33=3.所以an=a1+(n-1)d=3n.設(shè)等比數(shù)列bn-an的公比為q,由題意得q3=b4-a4b1-a1=20-124-3=8,解得q=2.所以bn-an=(b1-a1)qn-1=2n-1.從而bn=3n+2n-1