2013届数学高考一轮复习同步训练 文科 第30讲《等差数列》北师大版必修5 含答案
课时作业(三十) [第30讲 等差数列]
[时间:45分钟 分值:100分]
基础热身
1(等差数列{a}的前n项和为S,若S,4,S,20,则该数列的公差为( ) nn24
A(7 B(6 C(3 D(2
2([2011?介休一中月考] 等差数列{a}中,a,a,a,12,那么a,a,…,a,a,n3451267( )
A(21 B(28 C(32 D(35
3([2011?武汉模拟] 已知数列{a}是等差数列,若a,a,a,2π,则cos(a,a),( ) n15928
13A(, B(, 22
13C. D. 22*4([2011?天津卷] 已知{a}是等差数列,S为其前n项和,n?N.若a,16,S,20,则nn320S的值为________( 10
能力提升
5([2011?葫芦岛五校联考] 若等差数列{a}满足a,S,4,a,S,12,则a,S的值是n233547( )
A(20 B(36
C(24 D(72
6([2011?重庆三诊] 已知等差数列{a}满足a,a,a,2,则{a}的前15项和S,n3138n15( )
A(10 B(15
C(30 D(60
7([2011?南昌二模] 在等差数列{a}中,首项a,0,公差d?0,若a,a,a,…,a,n1k127则k,( )
A(21 B(22
C(23 D(24
,,1,,8([2011?郑州三模] 已知数列{a}中,a,2,a,1,且数列是等差数列,则an3711,a,1,n等于( )
21A(, B. 52
2C. D(5 3
9(已知数列{a}满足a,a,1(n?N),且a,a,a,18,则log(a,a,a)的值为,,nn1n2463579( )
A(,3 B(3
C(2 D(,2
10([2011?辽宁卷] S为等差数列{a}的前n项和,S,S,a,1,则a,________. nn2645
11(等差数列{a}中,若公差d,2,a,a,a,…,a,48,则a,a,a,…,a,n1472836930________.
12([2011?重庆卷] 在等差数列{a}中,a,a,37,则a,a,a,a,________. n372468
13([2011?惠州模拟] 已知等差数列{a}中,a,6,a,15,若b,a,则数列{b}的前n25n3nn9项和等于________(
14((10分)[2011?福建卷] 已知等差数列{a}中,a,1,a,,3. n13
(1)求数列{a}的通项公式; n
(2)若数列{a}的前k项和S,,35,求k的值( nk
15((13分)在数列{a}中,a,4,且对任意大于1的正整数n,点(a,a)在直线y,n1nn1,x,2上(
(1)求数列{a}的通项公式; n
(2)已知数列{b}的前n项和b,b,…,b,a,试比较a与b的大小( n12nnnn
难点突破 216((12分)数列{a}满足a,1,a,(n,n,λ)a(n,1,2,…),λ是常数( ,n1n1n
(1)当a,,1时,求λ及a的值; 23
(2)数列{a}是否可能为等差数列,若可能,求出它的通项公式;若不可能,说明理由( n
课时作业(三十)
【基础热身】
1(C [解析] S,2a,d,4,S,4a,6d,20,解得d,3.故选C. 2141
2(B [解析] 因为2a,a,a,所以3a,12,即a,4,所以a,a,…,a,a,7a4354412674,28.故选B.
2π4π13(A [解析] 由已知得a,,而a,a,2a,,所以cos(a,a),,.故选A. 528528332
4(110 [解析] 设等差数列的首项为a,公差为d,由
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
意得,1
a,a,2d,16,31,,,20, 解之得a,120×19S,20a,×d,20,20 1,,2
10×9d,,2,?S,10×20,×(,2),110. 102
【能力提升】
5(C [解析] 本题考查了等差数列的性质,因S,3a,得a,1,S,5a,得a,2,322533则a,3.又S,7a,则a,S,8a,24. 474474
,a,15,a1156(C [解析] 由a,a,a,2,得2a,a,2,所以a,2,所以S,,15a31388881582,30.故选C.
7×,7,1,d7(B [解析] 由已知等式得(k,1)d,,所以k,1,21,即k,22.故选B. 2
,,111111,,8(B [解析] 设,,4d,解得d,,所以,的公差为d,则有,a,,1a,1a,124a,1a,1n73113
1111,8d,即,.故选B. ,,,解得a11a,12,13211
9(B [解析] 因为{a}是等差数列,公差为1,且a,a,a,18,所以a,a,a,27,n246579所以所求值为3.故选B.
6×510(,1 [解析] 由S,S,得2a,d,6a,d解得4(a,3d),2d,0,即2a,d,2611142
0,所以a,(a,d),0,即a,,a,,1. 4454
11(88 [解析] a,a,a,…,a,a,a,a,…,a,20d,88. 3693014728
12(74 [解析] 由a,a,37,得(a,2d),(a,6d),37,即2a,8d,37.?a,a,a37111246,a,(a,d),(a,3d),(a,5d),(a,7d),2(2a,8d),74. 811111
,,a,a,d,6,a,3,211,,,,13(405 [解析] 由所以a,3,3(n,1),3n,b,a,9n,?nn3n ad,3,,,a,4d,15,,,51
9,81数列{b}的前9项和为S,×9,405. n92
14([解答] (1)设等差数列{a}的公差为d,则a,a,(n,1)d. nn1
由a,1,a,,3,可得1,2d,,3. 13
解得d,,2.
从而,a,1,(n,1)×(,2),3,2n. n
(2)由(1)可知a,3,2n. n
n[1,,3,2n,]2所以S,,2n,n. n22进而由S,,35可得2k,k,,35. k2即k,2k,35,0,解得k,7或k,,5. *又k?N,故k,7为所求(
15([解答] (1)因为点(a,a)在直线y,x,2上, ,nn1
所以a,a,2,即数列{a}是以a,2为首项,以d,2为公差的等差数列( ,nn1n1
所以a,2,2(n,1),2n, n
2所以a,4n. n22(2)方法一:因为b,b,…,b,a,所以当n?2时,b,a,a,4n,4(n,1),8n,12nnnnn1
,4,
当n,1时,b,a,4,满足上式(所以b,8n,4, 11n22所以a,b,4n,(8n,4),4(n,1)?0,所以a?b. nnnn方法二:由b,b,…,b,a得,a,b,a, ,12nnnnn124(n,1)?0,所以a?b. nn
【难点突破】 216([解答] (1)由于a,(n,n,λ)a(n,1,2,…),且a,1, ,n1n1所以当a,,1时,得,1,2,λ,故λ,3. 22从而a,(2,2,3)×(,1),,3. 3
(2)数列{a}不可能为等差数列(证明如下: n2由a,1,a,(n,n,λ)a得: ,1n1n
a,2,λ,a,(6,λ)(2,λ), 23
a,(12,λ)(6,λ)(2,λ)( 4
若存在λ,使{a}为等差数列,则a,a,a,a, n3221即(5,λ)(2,λ),1,λ,解得λ,3.
于是a,a,1,λ,,2,a,a,(11,λ)(6,λ)(2,λ),,24. 2143
这与{a}为等差数列矛盾(所以,对任意λ,{a}都不可能是等差数列( nn