（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練23 等差數(shù)列及其前n項(xiàng)和（含解析）新人教A版

考點(diǎn)規(guī)范練23等差數(shù)列及其前n項(xiàng)和一、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.272B.27C.54D.1082.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?()(注:尺是中國(guó)古代計(jì)量單位,1米=3尺)A.3B.4C.5D.63.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SnSn-1-Sn-1Sn=2SnSn-1(nN*,且n2),則a81等于()A.638B.639C.640D.6414.設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,且a1>0,a3+a10>0,a6a7<0,則滿足Sn>0的最大自然數(shù)n的值為()A.6B.7C.12D.135.若等差數(shù)列an的前n項(xiàng)和為Sn,且滿足a2+S3=4,a3+S5=12,則a4+S7的值是()A.20B.36C.24D.726.已知an為等差數(shù)列,Sn為其前n項(xiàng)和.若a2=2,S9=9,則a8=. 7.已知在數(shù)列an中,a1=1,a2=2,其前n項(xiàng)和為Sn.當(dāng)整數(shù)n2時(shí),Sn+1+Sn-1=2(Sn+S1)都成立,則S15=. 8.已知等差數(shù)列的前三項(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)bn=Snn,證明:數(shù)列bn是等差數(shù)列,并求其前n項(xiàng)和Tn.二、能力提升9.已知函數(shù)f(x)的圖象關(guān)于直線x=-1對(duì)稱,且f(x)在(-1,+)內(nèi)單調(diào),若數(shù)列an是公差不為0的等差數(shù)列,且f(a50)=f(a51),則數(shù)列an的前100項(xiàng)的和為()A.-200B.-100C.-50D.010.等差數(shù)列an的前n項(xiàng)和為Sn,已知a1=13,S3=S11,當(dāng)Sn最大時(shí),n的值是()A.5B.6C.7D.811.已知等差數(shù)列an的前n項(xiàng)和為Sn,a2<0,且1,a2,81成等比數(shù)列,a3+a7=-6.(1)求an的通項(xiàng)公式;(2)求Snn的前n項(xiàng)和Tn取得最小值時(shí)n的值.12.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a3·a4=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=Snn+c,求非零常數(shù)c.三、高考預(yù)測(cè)13.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.考點(diǎn)規(guī)范練23等差數(shù)列及其前n項(xiàng)和1.B解析S9=9(a1+a9)2=9(a2+a8)2=27.2.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,S30=302(a1+a30),即390=15(a1+21),解得a1=5.故選C.3.C解析由已知SnSn-1-Sn-1Sn=2SnSn-1,可得Sn-Sn-1=2,Sn是以1為首項(xiàng),2為公差的等差數(shù)列,故Sn=2n-1,Sn=(2n-1)2,a81=S81-S80=1612-1592=640,故選C.4.C解析因?yàn)閍1>0,a6a7<0,所以a6>0,a7<0,等差數(shù)列的公差小于零.又a3+a10=a1+a12>0,a1+a13=2a7<0,所以S12>0,S13<0,所以滿足Sn>0的最大自然數(shù)n的值為12.5.C解析由a2+S3=4及a3+S5=12,得4a1+4d=4,6a1+12d=12,解得a1=0,d=1,所以a4+S7=8a1+24d=24.故選C.6.0解析an為等差數(shù)列,Sn為其前n項(xiàng)和,a2=2,S9=9,a2=a1+d=2,S9=9a1+9×82d=9,解得d=-13,a1=73,a8=a1+7d=0.7.211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+2×14+14×132×2=211.8.(1)解設(shè)該等差數(shù)列為an,則a1=a,a2=4,a3=3a,由已知有a+3a=8,得a1=a=2,公差d=4-2=2,所以Sk=ka1+k(k-1)2·d=2k+k(k-1)2×2=k2+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)2=n(n+1),則bn=Snn=n+1,故bn+1-bn=(n+2)-(n+1)=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(2+n+1)2=n(n+3)2.9.B解析因?yàn)楹瘮?shù)f(x)的圖象關(guān)于直線x=-1對(duì)稱,函數(shù)f(x)在(-1,+)內(nèi)單調(diào),所以f(x)在(-,-1)內(nèi)也單調(diào),且數(shù)列an是公差不為0的等差數(shù)列.又f(a50)=f(a51),所以a50+a51=-2,所以S100=100(a1+a100)2=50(a50+a51)=-100.10.C解析(方法一)由S3=S11,得a4+a5+a11=0,根據(jù)等差數(shù)列的性質(zhì),可得a7+a8=0.根據(jù)首項(xiàng)等于13可推知這個(gè)數(shù)列為遞減數(shù)列,從而得到a7>0,a8<0,故n=7時(shí)Sn最大.(方法二)由S3=S11,可得3a1+3d=11a1+55d,把a(bǔ)1=13代入,得d=-2,故Sn=13n-n(n-1)=-n2+14n.根據(jù)二次函數(shù)的性質(zhì),知當(dāng)n=7時(shí)Sn最大.(方法三)由a1=13,S3=S11,知這個(gè)數(shù)列的公差不等于零,且這個(gè)數(shù)列的和是先遞增后遞減.根據(jù)公差不為零的等差數(shù)列的前n項(xiàng)和是關(guān)于n的二次函數(shù),以及二次函數(shù)圖象的對(duì)稱性,可得只有當(dāng)n=3+112=7時(shí),Sn取得最大值.11.解(1)a3+a7=-6=2a5,a5=-3.1,a2,81成等比數(shù)列,a22=1×81.又a2<0,a2=-9.等差數(shù)列an的公差d=a5-a25-2=-3-(-9)5-2=2.an=a2+(n-2)×2=2n-13.(2)Sn=n(-11+2n-13)2=n2-12n,Snn=n-12.由n-120,解得n12.因此,當(dāng)n=11或n=12時(shí),Snn的前n項(xiàng)和Tn取得最小值.12.解(1)數(shù)列an為等差數(shù)列,a3+a4=a2+a5=22.又a3·a4=117,a3,a4是方程x2-22x+117=0的兩實(shí)根.又公差d>0,a3<a4,a3=9,a4=13,a1+2d=9,a1+3d=13,a1=1,d=4.通項(xiàng)公式an=4n-3.(2)由(1)知a1=1,d=4,Sn=na1+n(n-1)2d=2n2-n=2n-142-18.當(dāng)n=1時(shí),Sn最小,最小值為S1=a1=1.(3)由(2)知Sn=2n2-n,bn=Snn+c=2n2-nn+c,b1=11+c,b2=62+c,b3=153+c.數(shù)列bn是等差數(shù)列,2b2=b1+b3,即62+c×2=11+c+153+c,2c2+c=0,c=-12(c=0舍去),故c=-12.13.解(1)a4=2a2,且a1,4,a4成等比數(shù)列,a1+3d=2(a1+d),a1·(a1+3d)=16,解得a1=2,d=2.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:20n116;n能夠被5整除,滿足條件的n組成等差數(shù)列bn,且b1=20,d=5,bn=115,項(xiàng)數(shù)為115-205+1=20.bn的所有項(xiàng)的和為S20=20×20+12×20×19×5=1350.又an=2n,即an=2bn,滿足條件的所有an的和為2S20=2×1350=2700.6