高考數(shù)學(xué)一輪復(fù)習(xí) 第六章 數(shù)列 考點(diǎn)規(guī)范練29 等差數(shù)列及其前n項(xiàng)和 文 新人教A版.doc

考點(diǎn)規(guī)范練29等差數(shù)列及其前n項(xiàng)和基礎(chǔ)鞏固1.已知sn為等差數(shù)列an的前n項(xiàng)和,a2+a8=6,則s9等于()a.b.27c.54d.1082.已知an是公差為1的等差數(shù)列,sn為an的前n項(xiàng)和.若s8=4s4,則a10=()a.b.c.10d.123.(2017陜西咸陽(yáng)二模)張丘建算經(jīng)卷上一題為“今有女善織,日益功疾,且從第二天起,每天比前一天多織相同量的布,現(xiàn)在一月(按30天計(jì))共織布390尺,最后一天織布21尺”,則該女第一天織布多少尺?()a.3b.4c.5d.64.已知在每項(xiàng)均大于零的數(shù)列an中,首項(xiàng)a1=1,且前n項(xiàng)和sn滿足sn-sn-1=2(nn*,且n2),則a81等于()a.638b.639c.640d.6415.已知數(shù)列an是等差數(shù)列,a1+a3+a5=105,a2+a4+a6=99,an的前n項(xiàng)和為sn,則使得sn達(dá)到最大的n是()a.18b.19c.20d.216.在等差數(shù)列an中,若是一個(gè)與n無(wú)關(guān)的常數(shù),則該常數(shù)的可能值的集合為()a.1b.c.d.7.已知等差數(shù)列an的前n項(xiàng)和為sn,且s10=10,s20=30,則s30=.8.已知在數(shù)列an中,a1=1,a2=2,當(dāng)整數(shù)n2時(shí),sn+1+sn-1=2(sn+s1)都成立,則s15=.9.若數(shù)列an的前n項(xiàng)和為sn,且滿足an+2snsn-1=0(n2),a1=.(1)求證:成等差數(shù)列;(2)求數(shù)列an的通項(xiàng)公式.10.在等差數(shù)列an中,a3+a4=4,a5+a7=6.(1)求an的通項(xiàng)公式;(2)設(shè)bn=an,求數(shù)列bn的前10項(xiàng)和,其中x表示不超過x的最大整數(shù),如0.9=0,2.6=2.能力提升11.若數(shù)列an滿足:a1=19,an+1=an-3(nn*),則當(dāng)數(shù)列an的前n項(xiàng)和數(shù)值最大時(shí),n的值為()a.6b.7c.8d.912.(2017四川廣元二診)設(shè)等差數(shù)列an的前n項(xiàng)和為sn,若sm-1=-2,sm=0,sm+1=3,其中m2,則nsn的最小值為()a.-3b.-5c.-6d.-913.數(shù)列an是等差數(shù)列,數(shù)列bn滿足bn=anan+1an+2(nn*),設(shè)sn為bn的前n項(xiàng)和.若a12=a50,則當(dāng)sn取得最大值時(shí),n的值等于.14.已知等差數(shù)列an的前n項(xiàng)和為sn,a20,且1,a2,81成等比數(shù)列,a3+a7=-6.(1)求an的通項(xiàng)公式;(2)求的前n項(xiàng)和tn取得最小值時(shí)n的值.15.已知公差大于零的等差數(shù)列an的前n項(xiàng)和為sn,且滿足a3a4=117,a2+a5=22.(1)求通項(xiàng)公式an;(2)求sn的最小值;(3)若數(shù)列bn是等差數(shù)列,且bn=,求非零常數(shù)c.高考預(yù)測(cè)16.已知各項(xiàng)均為正數(shù)的等差數(shù)列an滿足:a4=2a2,且a1,4,a4成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)求同時(shí)滿足下列條件的所有an的和:20n116;n能夠被5整除.答案：1.b解析:s9=27.2.b解析:公差d=1,s8=4s4,即2a1+7d=4a1+6d,解得a1=.a10=a1+9d=+9=.3.c解析:設(shè)第n天織布an尺,則數(shù)列an是等差數(shù)列,且s30=390,a30=21,則s30=(a1+a30),即390=15(a1+21),解得a1=5.故選c.4.c解析:由已知sn-sn-1=2,可得=2,是以1為首項(xiàng),2為公差的等差數(shù)列,故=2n-1,sn=(2n-1)2,a81=s81-s80=1612-1592=640,故選c.5.c解析:a1+a3+a5=105a3=35,a2+a4+a6=99a4=33,則an的公差d=33-35=-2,a1=a3-2d=39,sn=-n2+40n,因此當(dāng)sn取得最大值時(shí),n=20.6.b解析:特殊值驗(yàn)證法.若=1,則數(shù)列an是一個(gè)常數(shù)列,滿足題意;若,設(shè)等差數(shù)列的公差為d,則an=a2n=(an+nd),化簡(jiǎn),得an=nd,即a1+(n-1)d=nd,化簡(jiǎn),得a1=d,也滿足題意;若=0,則an=0,不符合題意.故選b.7.60解析:sn是等差數(shù)列an的前n項(xiàng)和,s10,s20-s10,s30-s20也成等差數(shù)列.2(s20-s10)=s10+(s30-s20).s30=60.8.211解析:由sn+1+sn-1=2(sn+s1)得(sn+1-sn)-(sn-sn-1)=2s1=2,即an+1-an=2(n2),數(shù)列an從第二項(xiàng)起構(gòu)成以2為首項(xiàng),2為公差的等差數(shù)列,則s15=1+214+2=211.9.(1)證明:當(dāng)n2時(shí),由an+2snsn-1=0,得sn-sn-1=-2snsn-1,所以=2.又=2,故是首項(xiàng)為2,公差為2的等差數(shù)列.(2)解:由(1)可得=2n,sn=.當(dāng)n2時(shí),an=sn-sn-1=-.當(dāng)n=1時(shí),a1=不適合上式.故an=10.解:(1)設(shè)數(shù)列an的公差為d,由題意有2a1+5d=4,a1+5d=3,解得a1=1,d=.所以an的通項(xiàng)公式為an=.(2)由(1)知,bn=.當(dāng)n=1,2,3時(shí),12,bn=1;當(dāng)n=4,5時(shí),23,bn=2;當(dāng)n=6,7,8時(shí),34,bn=3;當(dāng)n=9,10時(shí),40,得a1=-d,a12a5,即d0;當(dāng)n17時(shí),anb2b140b17b18,b15=a15a16a170,故s14s13s1,s14s15,s15s17s18.因?yàn)閍15=-d0,a18=d0,所以a15+a18=-d+d=d0,所以s16s14,故sn中s16最大.故答案為16.14.解:(1)a3+a7=-6=2a5,a5=-3.1,a2,81成等比數(shù)列,=181.又a20,a3a4,a3=9,a4=13,通項(xiàng)公式an=4n-3.(2)由(1)知a1=1,d=4,sn=na1+d=2n2-n=2.當(dāng)n=1時(shí),sn最小,最小值為s1=a1=1.(3)由(2)知sn=2n2-n,bn=,b1=,b2=,b3=.數(shù)列bn是等差數(shù)列,2b2=b1+b3,即2=,2c2+c=0,c=-(c=0舍去),故c=-.16.解:(1)a4=2a2,且a1,4,a4成等比數(shù)列,解得數(shù)列an的通項(xiàng)公式為an=a1+(n-1)d=2+2(n-1)=2n.(2)n同時(shí)滿足