高考數(shù)學(xué)大二輪復(fù)習(xí) 能力升級(jí)練（七）等差數(shù)列與等比數(shù)列 理-人教版高三數(shù)學(xué)試題

能力升級(jí)練(七)等差數(shù)列與等比數(shù)列一、選擇題1.已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=8,則a100=()A.100 B.99 C.98 D.97解析設(shè)等差數(shù)列{an}的公差為d,由已知,得9a1所以a100=a1+99d=-1+99=98.答案C2.公比不為1的等比數(shù)列{an}滿足a5a6+a4a7=18,若a1am=9,則m的值為()A.8 B.9 C.10 D.11解析由題意得,2a5a6=18,a5a6=9,∴a1am=a5a6=9,∴m=10.答案C3.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a4與a14的等比中項(xiàng)為22,則2a7+a11的最小值為()A.16 B.8 C.22 D.4解析因?yàn)閍4與a14的等比中項(xiàng)為22,所以a4·a14=a7·a11=(22)2=8,所以2a7+a11≥22a7a11=當(dāng)且僅當(dāng)a7=2,a11=4時(shí),取等號(hào),所以2a7+a11的最小值為8.答案B4.(2019山東淄博調(diào)研)設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,若a6a5=911A.1 B.-1 C.2 D.1解析由于S11S9答案A5.(2019中原名校聯(lián)考)若數(shù)列{an}滿足1an+1-1an=d(n∈N*,d為常數(shù)),則稱數(shù)列{an}為調(diào)和數(shù)列,已知數(shù)列1xn為調(diào)和數(shù)列,且x1+x2+…+x20=A.10 B.20 C.30 D.40解析依題意,11xn+1-11xn=xn+1-xn=d,∴{xn}是等差數(shù)列.又x1+x2∴x1+x20=20,從而x5+x16=x1+x20=20.答案B6.(2019北京海淀質(zhì)檢)中國古詩詞中,有一道“八子分綿”的數(shù)學(xué)名題:“九百九十六斤綿,贈(zèng)分八子作盤纏,次第每人多十七,要將第八數(shù)來言”.題意是:把996斤綿分給8個(gè)兒子作盤纏,按照年齡從大到小的順序依次分綿,年齡小的比年齡大的多17斤綿,那么第8個(gè)兒子分到的綿是()A.174斤 B.184斤 C.191斤 D.201斤解析用a1,a2,…,a8表示8個(gè)兒子按照年齡從大到小得到的綿數(shù),由題意得數(shù)列a1,a2,…,a8是公差為17的等差數(shù)列,且這8項(xiàng)的和為996,∴8a1+8×72×17=996,解得a1=65.∴a8=65+7×17=184,即第8個(gè)兒子分到的綿是184斤答案B7.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a1=9,S99-S55=-4,則SA.4 B.5C.6 D.4或5解析由{an}為等差數(shù)列,得S99-S55=a5-a3=2d=-4,即d=-2,由于a1=9,所以an=-2n+11,令an=-2n+11<0,得n>112答案B8.(2019上海崇明模擬)已知公比q≠1的等比數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,S3=3a3,則S5=()A.1 B.5C.3148 D.解析由題意得a1(1-q3)1-q=3a1q2,解得答案D9.(2019廣東深圳一模)已知等比數(shù)列{an}的前n項(xiàng)和Sn=a·3n-1+b,則ab=(A.-3 B.-1C.1 D.3解析∵等比數(shù)列{an}的前n項(xiàng)和Sn=a·3n-1+b,∴a1=S1=a+b,a2=S2-S1=3a+b-a-b=2a,a3=S3-S2=9a+b-3a-b=6a,∵等比數(shù)列{an}中,a22=a1a3,∴(2a)2=(a+b)×6a,解得ab答案A二、填空題10.已知等差數(shù)列{an}的公差為2,項(xiàng)數(shù)是偶數(shù),所有奇數(shù)項(xiàng)之和為15,所有偶數(shù)項(xiàng)之和為25,則這個(gè)數(shù)列的項(xiàng)數(shù)為.

解析設(shè)項(xiàng)數(shù)為2n,則由S偶-S奇=nd得,25-15=2n,解得n=5,故這個(gè)數(shù)列的項(xiàng)數(shù)為10.答案1011.設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,S10=16,S100-S90=24,則S100=.

解析依題意,S10,S20-S10,S30-S20,…,S100-S90依次成等差數(shù)列,設(shè)該等差數(shù)列的公差為d.又S10=16,S100-S90=24,因此S100-S90=24=16+(10-1)d=16+9d,解得d=89,因此S100=10S10+10×92d=10×16+10×92答案20012.等比數(shù)列{an}中,各項(xiàng)都是正數(shù),且a1,12a3,2a2成等差數(shù)列,則a13+解析設(shè){an}的公比為q.由題意,得a1+2a2=a3,則a1(1+2q)=a1q2,q2-2q-1=0,所以q=1+2,或q=1-2(舍).則a13+a答案2-113.(2018江蘇南京模擬)已知數(shù)列{an}中,a1=2,且an+12an=4(an+1-an)(n∈N*),則其前9項(xiàng)的和解析由an+12an=4(an+1-an),得an+12-4an+∴(an+1-2an)2=0,an+1an=2,∴數(shù)列{an}是首項(xiàng)a1=2,公比為2的等比數(shù)列,∴S9=答案1022三、解答題14.已知等差數(shù)列{an}的前三項(xiàng)依次為a,4,3a,前n項(xiàng)和為Sn,且Sk=110.(1)求a及k的值;(2)設(shè)數(shù)列{bn}的通項(xiàng)公式bn=Snn,證明:數(shù)列{bn}是等差數(shù)列,并求其前n項(xiàng)和T(1)解由題意,得a1=a,a2=4,a3=3a,∵a+3a=8,∴a1=a=2,公差d=4-2=2,∴Sk=ka1+k(k-1)2·d=2k+k由Sk=110,得k2+k-110=0,解得k=10或k=-11(舍去),故a=2,k=10.(2)證明由(1)得Sn=n(2+2n)則bn=Snn=n+1,故bn+1-bn=(n+2)-(n+1)即數(shù)列{bn}是首項(xiàng)為2,公差為1的等差數(shù)列,所以Tn=n(15.已知數(shù)列{an}中,點(diǎn)(an,an+1)在直線y=x+2上,且首項(xiàng)a1=1.(1)求數(shù)列{an}的通項(xiàng)公式;(2)數(shù)列{an}的前n項(xiàng)和為Sn,等比數(shù)列{bn}中,b1=a1,b2=a2,數(shù)列{bn}的前n項(xiàng)和為Tn,請寫出適合條件Tn≤Sn的所有n的值.解(1)根據(jù)已知a1=1,an+1=an+2,即an+1-an=2=d,所以數(shù)列{an}是一