专题09等差数列、等比数列1.已知等差数列{an}的前n项和为Sn,若a3+a5=8,则S7=()A.28B.32C.56D.24解析:S7===28.故选A.答案:A2.等比数列{an}的前n项和为Sn,若2S4=S5+S6,则数列{an}的公比q的值为()A.-2或1B.-1或2C.-2D.1答案:C3.设等差数列{an}的前n项和为Sn,a1>0且=,则当Sn取最大值时,n的值为()A.9B.10C.11D.12解析:由题意,不妨设a6=9t,a5=11t,则公差d=-2t,其中t>0,因此a10=t,a11=-t,即当n=10时,Sn取得最大值.答案:B4.在各项均为正数的等比数列{an}中,若am+1·am-1=2am(m≥2),数列{an}的前n项积为Tn,若T2m-1=512,则m的值为()A.4B.5C.6D.7解析:由等比数列的性质可知am+1·am-1=a=2am(m≥2),∴am=2,即数列{an}为常数列,an=2,∴T2m-1=22m-1=512=29,即2m-1=9,所以m=5.答案:B5.已知等比数列{an}的各项都是正数,且3a1,a3,2a2成等差数列,则=()A.6B.7C.8D.9解析:∴3a1,a3,2a2成等差数列,∴a3=3a1+2a2,∴q2-2q-3=0,∴q=3或q=-1(舍去).∴===q2=32=9.答案:D6.等比数列{an}的前n项和为Sn,若a1+a2+a3+a4=1,a5+a6+a7+a8=2,Sn=15,则项数n为()A.12B.14C.15D.16答案:D7.已知各项均为正数的等比数列{an}中,a1a2a3=5,a7a8a9=10,则a4a5a6=()A.5B.7C.6D.4答案:A解析:由题意知a1a2a3,a4a5a6,a7a8a9三数成等比数列,所以(a4a5a6)2=(a1a2a3)·(a7a8a9)=50.又an>0,∴a4a5a6=5,故选A.8.已知Sn是等比数列{an}的前n项和,若存在m∈N*,满足=9,=,则数列{an}的公比为()A.-2B.2C.-3D.3答案:B解析:设公比为q,若q=1,则=2,与题中条件矛盾,故q≠1. ==qm+1=9,∴qm=8.∴==qm=8=,∴m=3,∴q3=8,∴q=2.9.已知等比数列{an}中,a2=1,则其前3项的和S3的取值范围是()A.(-∞,-1]B.(-∞,0)∪(1,+∞)C.[3,+∞)D.(-∞,-1]∪[3,+∞)答案:D10.数列{an}满足a1=2且对任意的m,n∈N*,都有=an,则a3=________;{an}的前n项和Sn=________.【解析】 =an,∴an+m=an·am,∴a3=a1+2=a1·a2=a1·a1·a1=23=8.令m=1,则有an+1=an·a1=2an,∴数列{an}是首项为a1=2,公比为q=2的等比数列,∴Sn==2n+1-2.【答案】82n+1-211.已知数列{an}满足a1=a2=1,an+2=an+1+an(n∈N*).若存在正实数λ使得数列{an+1+λan}为等比数列,则λ=________.【解析】由题意可知,an+2+λan+1=(1+λ)an+1+an=(1+λ),∴=λ,解得λ=或λ=(舍), a1=a2=1,∴a3=2,易验证当n=1时满足题意.故λ=.【答案】12.各项均不为零的等差数列{an}中,a1=2,若a-an-1-an+1=0(n∈N*,n≥2),则S2016=________.解析:由于a-an-1-an+1=0(n∈N*,n≥2),即a-2an=0,∴an=2,n≥2,又a1=2,∴an=2,n∈N*,故S2016=4032.答案:403213.设数列{an}的前n项和为Sn.若S2=4,an+1=2Sn+1,n∈N*,则a1=________,S5=________.答案:112114.已知数列{an}的各项均为正数,Sn为其前n项和,且对任意n∈N*,均有an,Sn,a成等差数列,则an=________.解析: an,Sn,a成等差数列,∴2Sn=an+a.当n=1时,2a1=2S1=a1+a.又a1>0,∴a1=1.当n≥2时,2an=2(Sn-Sn-1)=an+a-an-1-a,∴(a-a)-(an+an-1)=0,∴(an+an-1)(an-an-1)-(an+an-1)=0,又an+an-1>0,∴an-an-1=1,∴{an}是以1为首项,1为公差的等差数列,∴an=n(n∈N*).答案:n15.已知等差数列{an}满足a3=2,前3项和S3=.(1)求{an}的通项公式;(2)设等比数列{bn}满足b1=a1,b4=a15,求{bn}的前n项和Tn.16.设数列{an}的前n项和为Sn,n∈N*.已知a1=1,a2=,a3=,且当n≥2时,4Sn+2+5Sn=8Sn+1+Sn-1.(1)求a4的值;(2)证明:为等比数列;(3)求数列{an}的通项公式.(1)解:当n=2时,4S4+5S2=8S3+S1,即4(a1+a2+a3+a4)+5(a1+a2)=8(a1+a2+a3)+a1,整理得a4=,又a2=,a3=,所以a4=.(2)证明:当n≥2时,有4Sn+2+5Sn=8Sn+1+Sn-1,即4Sn+2+4Sn+Sn=4Sn+1+4S...