贵州2014职高对口升学数学复习专
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
:数列01
贵州2014职高对口升学数学复习专题:数列01
一、选择题
21 (已知各项不为0的等差数列满足,数列是等比数列,且{}a{}b222aaa,,nn7122
,则= ( ) ba,bb7759
A(16 B(8 C(4 D(2 【
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
】A
2 (已知数列是正项等比数列,若,则数列的通项公式为 ( ) ,,,,aa,2,2a,a,16ann234
n,22,nn,1nA( B( C( D( 2222
【答案】C
3 (一个等差数列第5项,则有 ( ) aaaa,,,,10,3且6123
A( B( ad,,,2,3ad,,,2,311
C(ad,,,3,2 D(ad,,,3,2 23
【答案】A
aaa,,4aa,,10S,4 (已知等差数列满足,,则它的前10项的和 ( ) ,,n243510A(138 B(135 C(95 D(23 【答案】C
5 (已知等差数列{a}的前n项和为S,S=-18,S=-52,{b}为等比数列,且b =a,b=a,则nn913n5577b的值为 ( ) 15
A(64 B(128 C(-64 D(-128 【答案】C
,,aa,a,a6 (已知等比数列中,公比,若a,4,则最值情况为 ( ) q,0n1232
A(最小值 B(最大值 C(最小值 D(最大值 ,4,41212【答案】B
S{}aa,1SS,,24nd,2k,nn1kk,27 (设为等差数列的前项和,若,公差,,则 ( )
A(8 B(7 C(6 D(5 【答案】D
aaaaaaaaaa8 (已知各项均为正数的等比数列{},=5,=10,则= ( ) n123789456
5242A( B(7 C(6 D(
【答案】A
9 (等差数列的公差为2,若a,a,a成等比数列,则a= ( ) {a}1342n
A(-6 B(-8 C(8 D(6
【答案】A
10(已知等差数列的前项和为,且满足,则的值是 ( ) ,,naSS,5,tanann158
333,3A( B( C( D( ,33
【答案】A
11(已知{a}为等差数列,其前n项和为S,若a=6,S=12,则S等于 ( ) nn3312
A(288 B(90 C(156 D(126
【答案】C
12(已知各项均为正数的等比数列满足,若存在两项使得{}aaaa,,2aa,n765mn
14的最小值为 ( ) aaa,,则4,mn1mn
359A( B( C( D(不存在 243
【答案】A
,,1anSaS,5,15,,13(已知等差数列的前项和为,则数列的前100项和为 ( ) ,,,,nn55aann,1,,
1009999101A( B( C( D( 101100100101
【答案】A
14(已知数列的前n項和满足:,且,那么a为 ( ) 5
A(1 B(9 C(1O D(55
【答案】A
S6{}a80aa,,15(等比数列中,,则= ( ) n23s2
A(-10 B(10 C(20 D(21
【答案】D
aaaa,,,12aaa,,,,...16(如果等差数列中,,那么 ( ) ,,n345127
A(14 B(21 C(28 D(35
【答案】C
anSaSS,,0,S17(2012年5月7日 ,已知等差数列的前项和为,若,则当取得最小,,nn1610n
值时,的值为 ( ) n
A(6 B(7 C(8 D(10 【答案】 解:法一:由
2 S,S,S,S,a,a,a,a,0,a,a,0,d,,a,15所以可知该数列为单调递增数列,由可知,,故最小 a,a,0a,0,aS89898法二:若公差,则在上递减,有,与已知不符合 d,0SS,Sn,Nn610,
6,10所以,由可知这个关于的二次
函
关于工期滞后的函关于工程严重滞后的函关于工程进度滞后的回复函关于征求同志党风廉政意见的函关于征求廉洁自律情况的复函
数的对称轴为,所nd,0S,SSn,,8610n2以最小.选答案C S8
aa,189a18(已知等比数列中,各项都是正数,且,,成等差数列,则等于 ( ) a2aa,,n1232aa,67A( B( C( (D) 322,12,12,322,【答案】A
a19(等差数列的公差为2,若成等比数列,则a,aaa,,,,n2134 ( )
,8,6 B( C(8 D(6 A(
【答案】B
二、填空题
Snnan,,,52S20(已知数列的通项,其前项和为,则___________. lim,nn2?,nn
5【答案】 ,2
{}aS2S3S{}a21(等比数列的前n项和为,已知S,,成等差数列,则的公比为nn23n1
______.
n,1{}aS2S3SSaaq,【答案】等比数列的前n项和为,已知,,成等差数列,,nn231n1
1243SSS,,{}a4()3()aaqaaaqaq,,,,,又,即,解得的公比. q,213n1111113
S9n{}aSa,5a22(设等差数列的前项和为,.若,则=__________. n53mS5
【答案】9
anSS,72aaa,,23(设等差数列的前项和为,若,则=__________. ,,nn9249
?aS,72?,Sa9,a,8【答案】解: 是等差数列,由,得 ,,n9955
. aaaaaaaaaa,,,,,,,,,,()()324?2492945645
,a24(数列中,已知,,,则_______. a,1a,2a,aaanN,,,(),,n12nnn127,,【答案】1
三、解答题
11725(数列的前n项和. bbbTb,,,且为,,{}nnnn,11242
1(1)求证:数列是等比数列,并求的通项公式; {}bb,{}nn2
12k*(2)如果对任意恒成立,求实数k的取值{}bnNn,,,,27不等式n(122),,nTn范围.
11111*【答案】解: (1) 对任意,都有,所以 n,Nbb,,bb,,,()nn,1nn,124222
111则成等比数列,首项为,公比为 b,b,,3{}n1222
1111n,1n,1所以, b,,,b,,,3()3()nn2222
aSaSa,,,1,4226(设数列的前 n项和为,已知 ,,nn11nn,
ba2a{b},,,证明数列是等比数列(1) 设 nn1nn,
(2) 求数列{a}的通项公式 n
【答案】解:(I)由及,有a,1,Sa,,421nn,1
aaa,,,42,aabaa,,,?,,,325,2312121121
由,(((? 则当时,有(((((? n,2Sa,,42Sa,,42nn,1nn,1?,?得 aaaaaaa,,?,,,44,22(2)nnnnnnn,,,,1111
又,是首项,公比为,的等比数列( ?baa,,2?,bb2?{}bb,3nnn,1nn,1n1
aa3n,1nn,1(II)由(I)可得, baa,,,,232?,,nnn,1nn,1224
a13n数列是首项为,公差为的等比数列( {}?n224
a1331n,2n, an,,,(31)2,,,,,(1)nn?nn22444