1、见学生用书第248页一、选择题1(2011广东三校模拟)已知数列an是等差数列,且a1a3a52,则cos a3()A.BC.D2已知等差数列an的前n项和为Sn,a415,S555,则过点P(3,a3),Q(10,a10)直线的斜率为()A4 B28 C4 D143已知数列an则a1a2a3a4a99a100()A4 800 B4 900 C5 000 D5 1004已知等差数列an与bn的前n项和分别为Sn与Tn,且,则等于()A. B. C. D.5(2011长沙模拟)若an是等差数列,首项a10,a2 003a2 0040,a2 003a2 0040,则使数列an的前n项和Sn0成立的
2、最大自然数n是()A4 005 B4 006 C4 007 D4 008二、填空题6设等差数列an的前n项和为Sn,若a6S312,则an的通项公式an_. 7等差数列an的前n项和为Sn,且6S55S35,则a4_.8(2011朝阳模拟)各项均不为零的等差数列an中,若aan1an10(nN*,n2),则S2 012等于_三、解答题9在等差数列an中,已知a2a7a1212,a2a7a1228,求数列an的通项公式10已知数列an中a18,a42,且满足an2an2an1.(1)求数列an的通项公式;(2)设Sn是数列|an|的前n项和,求Sn.11(2011东城模拟)设数列an的前n项和为
3、Sn,已知a11,Snnann(n1)(n1,2,3,)(1)求证:数列an为等差数列,并写出an关于n的表达式;(2)若数列的前n项和为Tn,问满足Tn的最小正整数n是多少?答案及解析1【解】a1a52a3,a1a3a53a32,a3.cos a3cos .【答案】D2【解】S55a355,a311,公差da4a315114,直线PQ的斜率k4.【答案】A3【解】由题意得a1a2a3a4a99a1000224498981002(24698)10021005 000.【答案】C4【解】.【答案】B5【解】由a10,a2 003a2 0040,a2 003a2 0040得an是递减的等差数列,a
4、2 0030,a2 0040,又a2 003a2 004a1a4 0060,a1a4 0072a2 0040,S4 0060,S4 0070,最大自然数n是4 006.【答案】B6【解】由题意得解得ana1(n1)d2n.【答案】2n7【解】6S55S35,6(5a110d)5(3a13d)5,a13d,即a4.【答案】8【解】an1an12anaan1an1a2an0,解得an2或an0(舍)S2 01222 0124 024.【答案】4 0249【解】由a2a7a1212得a74.又a2a7a1228,(a75d)(a75d)a728,d2,d或d.d时,ana7(n7)d4(n7)n;d
5、时,ana7(n7)d4(n7)n.10【解】(1)由2an1an2an可得an是等差数列,且公差d2.ana1(n1)d2n10.(2)令an0得n5.即当n5时,an0,n6时,an0当n5时,Sn|a1|a2|an|a1a2ann29n当n6时,Sn|a1|a2|an|a1a2a5(a6a7an)(a1a2an)2(a1a2a5)(n29n)2(5245)n29n40,11【证明】(1)当n2时,anSnSn1nan(n1)an12(n1),得anan12(n2,3,4,)所以数列an是以a11为首项,2为公差的等差数列所以an2n1.(2)Tn()()()(1).由Tn,得n,满足Tn的最小正整数为12.