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1、新版數(shù)學(xué)北師大版精品資料學(xué)業(yè)水平訓(xùn)練1在等差數(shù)列an中,an0,且a1a2a1030,則a5a6()A3B6C9 D36解析:選B.數(shù)列an是等差數(shù)列,且an0,a1a2a105(a5a6)30,a5a66.2(2014·臨清高二檢測(cè))已知等差數(shù)列an中,a2a46,則a1a2a3a4a5()A30 B15C5 D10解析:選B.數(shù)列an為等差數(shù)列a1a2a3a4a5(a2a4)×615.3(2014·東北育才學(xué)校質(zhì)檢)在等差數(shù)列an中,若a1,a2 015為方程x210x160的兩根,則a2a1 008a2 014()A10 B15C20 D40解析:選B.a1
2、,a2 015為方程x210x160的兩個(gè)根a1a2 0152a1 00810.a1 0085,a2a1 008a2 0143a1 0083×515.4設(shè)an,bn都是等差數(shù)列,且a125,b175,a2b2100,則a37b37()A0 B37C100 D37解析:選C.設(shè)cnanbn,由于an,bn都是等差數(shù)列,則cn也是等差數(shù)列,且的公差dc2c10.c37100.5已知等差數(shù)列an的公差為d(d0),且a3a6a10a1332,若am8,則m等于()A8 B4C6 D12解析:選A.因?yàn)閍3a6a10a134a832,所以a88,即m8.6(2014·泰安高二檢測(cè))
3、在等差數(shù)列an中,a3,a10是方程x23x50的根,則a5a8_解析:由已知得a3a103,又?jǐn)?shù)列an為等差數(shù)列,a5a8a3a103.答案:37(2014·河北省石家莊市月考)在等差數(shù)列an中,若a3a5a7a9a11100,則3a9a13的值為_解析:由等差數(shù)列的性質(zhì)可知,a3a5a7a9a11(a3a11)(a5a9)a75a7100,a720.又3a9a132a9a9a13(a5a13)a9a13a5a92a740.答案:408已知數(shù)列an滿足a11,若點(diǎn)(,)在直線xy10上,則an_解析:由題設(shè)可得10,即1,所以數(shù)列是以1為公差的等差數(shù)列,且首項(xiàng)為1,故通項(xiàng)公式n,所
4、以ann2.答案:n29在等差數(shù)列an中:(1)若a3a9,求a6;(2)若a2a3a10a1148,求a6a7.解:在等差數(shù)列an中:(1)a3a92a6,a6.(2)a6a7a3a10a2a11,且a2a3a10a1148,2(a6a7)48,a6a724.10如果有窮數(shù)列a1,a2,am(m為正整數(shù))滿足條件:a1am,a2am1,ama1,那么稱其為“對(duì)稱”數(shù)列例如數(shù)列1,2,5,2,1與數(shù)列8,4,2,4,8都是“對(duì)稱”數(shù)列已知在21項(xiàng)的“對(duì)稱”數(shù)列cn中,c11,c12,c21是以1為首項(xiàng),2為公差的等差數(shù)列,求c2的值解:c11,c12,c21是以1為首項(xiàng),2為公差的等差數(shù)列,c
5、20c119d19×219,又cn為21項(xiàng)的對(duì)稱數(shù)列,c2c2019.高考水平訓(xùn)練1設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,若a11,公差d2,Sk2Sk24,則k()A8 B7C6 D5解析:選D.Sk2Skak1ak2a1kda1(k1)d2a1(2k1)d2×1(2k1)×24k424,k5.2(2014·銅陵調(diào)研)在等差數(shù)列an中,若a7m,a14n,則a21_解析:a7、a14、a21成等差數(shù)列,a7a212a14,a212a14a72nm.答案:2nm3(2014·北京東城區(qū)綜合練習(xí))已知f(x)是定義在R上不恒為零的函數(shù),對(duì)于任意的x,y
6、R,都有f(x·y)xf(y)yf(x)成立數(shù)列an滿足anf(2n)(nN)且a12,求數(shù)列an的通項(xiàng)公式解:令x2,y2n1,則f(x·y)f(2n)2f(2n1)2n1f(2),即f(2n)2f(2n1)2n1a1,即an2an12n,1,所以數(shù)列為以1為首項(xiàng),1為公差的等差數(shù)列,所以n.由此可得ann·2n.4在數(shù)列an中,a11,3anan1anan10(n2,nN)(1)求證:數(shù)列是等差數(shù)列;(2)求數(shù)列an的通項(xiàng)公式;(3)若an對(duì)任意n2的整數(shù)恒成立,求實(shí)數(shù)的取值范圍解:(1)證明:由3anan1anan10,得3(n2)又a11,數(shù)列是以1為首項(xiàng),3為公差的等差數(shù)列(2)由(1)可得13(n1)3n2,an.(3)an對(duì)任意n2的整數(shù)恒成立,即3n1對(duì)n2的整數(shù)恒成立整理,得,令cn,cn1cn.n2,cn1cn0,即數(shù)列cn為單調(diào)遞增數(shù)列,c2最小又c2,的取值范圍為(,