专题强化训练13

专题强化训练13
专题强化训练13

专题强化训练(十三)

一、选择题

1.(2020·江西红色七校第一次联考)已知等差数列{a n }的前n 项和S n 满足S 8-S 3=45,则a 6的值是( )

A .3

B .5

C .7

D .9

[解析] 解法一:设等差数列{a n }的公差为d ,则S 8-S 3=8a 1+

8×72×d -? ???

?

3a 1+3×22×d =5a 1+25d =45,即a 1+5d =9,即a 6=9,

故选D.

解法二:因为S 8-S 3=a 4+a 5+a 6+a 7+a 8=5a 6=45,所以a 6=9,故选D.

[答案] D

2.(2020·广东珠海模拟)已知等比数列{a n }的前n 项和S n ,且S 4

=15,a 2+a 4=10,则a 2=( )

A .1

B .-2

C .2

D .-1

[解析] 设等比数列{a n }的公比为q (q ≠0).

∵??

?

S 4=15,a 2+a 4=10,

∴??

?

a 1+a 1q +a 1q 2+a 1q 3=15,a 1q +a 1q 3=10,

解得a 1=1,q

=2,∴a 2=1×2=2.故选C.

[答案] C

3.(2020·大同高三调研)若等差数列{a n }的前n 项和S n 有最大值,且a 11

a 10

<-1,则S n 取正值时项数n 的最大值为( )

A .15

B .17

C .19

D .21

[解析] 由等差数列{a n }的前n 项和S n 有最大值,且a 11

a 10

<-1,可

知等差数列{a n }的公差d <0,a 10>0,a 11<0,且a 11<-a 10,则a 10+a 11<0.由a 10>0,得2a 10=a 1+a 19>0,所以S 19>0,由a 10+a 11<0,得a 1+a 20=a 10+a 11<0,所以S 20<0,所以S n 取正值时项数n 的最大值为19,故选C.

[答案] C

4.(2020·山东青岛一模)设{a n }是等差数列.下列结论中正确的是( )

A .若a 1+a 2>0,则a 2+a 3>0

B .若a 1+a 3<0,则a 1+a 2<0

C .若0a 1a 3

D .若a 1<0,则(a 2-a 1)(a 2-a 3)>0

[解析] 若{a n }是递减的等差数列,则选项A ,B 都不一定正确.若{a n }为公差为0的等差数列,则选项D 不正确.对于C 选项,由0

2>a 1a 3,选项C 正确.故选C.

[答案] C

5.(2020·江苏徐州期中)在公比q 为整数的等比数列{a n }中,S n

是数列{a n }的前n 项和.若a 1·a 4=32,a 2+a 3=12,则下列说法错误的是( )

A .q =2

B .数列{S n +2}是等比数列

C .S 8=510

D .数列{log 2a n }是公差为2的等差数列

[解析] 因为数列{a n }为等比数列,又a 1·a 4=32, 所以a 2·a 3=32,又a 2+a 3=12,

所以????? a 2=4,

a 3

=8,

q =2,

或???

?? a 2=8,a 3

=4,q =12.

又公比q 为整数,则????

?

a 2=4,

a 3

=8,

q =2,

即a n =2n ,S n =2×(1-2n )

1-2

=2n +1-2.

对于选项A ,由上可得q =2,即选项A 正确;

对于选项B ,S n +2=2n +1,S n +1+2S n +2=2n +2

2n +

1=2,则数列{S n +2}是

等比数列,即选项B 正确;

对于选项C ,S 8=29-2=510,即选项C 正确;

对于选项D ,log 2a n +1-log 2a n =(n +1)-n =1,即数列{log 2a n }是公差为1的等差数列,即选项D 错误.

[答案] D

6.(2020·郑州二中期末)已知等差数列{a n }的公差d ≠0,且a 1,a 3,a 13成等比数列,若a 1=1,S n 是数列{a n }的前n 项的和,则

2S n +16a n +3(n ∈N *)的最小值为( )

A .4

B .3

C .23-2

D.92

[解析] ∵a 1=1,a 1、a 3、a 13成等比数列, ∴(1+2d )2=1+12d .得d =2或d =0(舍去) ∴a n =2n -1,

∴S n =n (1+2n -1)

2=n 2

, ∴2S n +16a n +3=2n 2+162n +2

.令t =n +1, 则2S n +16a n +3=t +9t -2≥6-2=4当且仅当t =3, 即n =2时等号成立,∴2S n +16

a n +3的最小值为4.故选A.

[答案] A 二、填空题

7.(2020·福建四地六校联考)已知等差数列{a n }中,a 3=π

4,则cos(a 1+a 2+a 6)=________.

[解析] ∵在等差数列{a n }中,a 1+a 2+a 6=a 2+a 3+a 4=3a 3=34π,∴cos(a 1+a 2+a 6)=cos 34π=-22.

[答案] -2

2

8.(2020·浙江嘉兴教学测试)已知{a n }是公差为-2的等差数列,S n 为其前n 项和.若a 2+1,a 5+1,a 7+1成等比数列,则a 1=________,当n =________时,S n 取得最大值.

[解析] 因为a 2+1,a 5+1,a 7+1成等比数列,

所以(a 5+1)2=(a 2+1)(a 7+1). 又{a n }是公差为-2的等差数列, 所以(a 1-8+1)2=(a 1-2+1)(a 1-12+1), 即(a 1-7)2=(a 1-1)(a 1-11),解得a 1=19,

所以S n =19n -n (n -1)=-n 2+20n =-(n -10)2+100. 因此,当n =10时,S n 取得最大值. [答案] 19 10

9.(2020·哈尔滨模拟)已知数列{a n }的前n 项和为S n ,S 1=6,S 2

=4,S n >0,且S 2n ,S 2n -1,S 2n +2成等比数列,S 2n -1,S 2n +2,S 2n +1成等差数列,则a 2020=________.

[解析]

由题意得??

?

S 2

2n -1=S 2n S 2n +2,

2S 2n +2=S 2n -1+S 2n +1.

∵S n >0,∴2S 2n +2=S 2n S 2n +2+S 2n +2S 2n +4,

即2

S 2n +2=S 2n +

S 2n +4(n ∈N *),

故{S 2n }是等差数列. 又由S 1=6,S 2=4,得S 4=9,

∴{S 2n }是首项为2,公差为1的等差数列. ∴S 2n =n +1,即S 2n =(n +1)2, 故S 2n -1=

S 2n S 2n +2=(n +1)·(n +2),

故S 2020=10112,S 2019=1011×1012, 故a 2020=S 2020-S 2019=-1011. [答案] -1011 三、解答题

10.(2020·成都测试)已知等差数列{a n }前三项的和为-9,前三项的积为-15.

(1)求等差数列{a n }的通项公式;

(2)若{a n }为递增数列,求数列{|a n |}的前n 项和S n .

[解] (1)设数列{a n }的公差为d ,则依题意得a 2=-3,则a 1=-3-d ,a 3=-3+d ,

∴(-3-d )(-3)(-3+d )=-15,解得d 2=4,d =±2, ∴等差数列{a n }的通项公式为a n =-2n +1或a n =2n -7.

(2)由题意及(1)得a n =2n -7,∴|a n |=??

? 7-2n ,n ≤3,2n -7,n ≥4,

①当n ≤3时,S n =-(a 1+a 2+…+a n )=5+(7-2n )

2n =6n -n 2

; ②当n ≥4时,S n =-a 1-a 2-a 3+a 4+…+a n =-2(a 1+a 2+a 3)+(a 1+a 2+…+a n )=18-6n +n 2.

综上,数列{|a n |}的前n 项和S n =??

?

-n 2+6n ,n ≤3,n 2-6n +18,n ≥4.

11.(2020·广西南宁调研)已知正项数列{a n }的前n 项和为S n ,a 1=1,S 2

n =a 2n +1-λS n +1,其中λ为常数.

(1)证明:S n +1=2S n +λ;

(2)是否存在实数λ,使得数列{a n }为等比数列?若存在,求出λ;若不存在,请说明理由.

[解] (1)证明:∵a n +1=S n +1-S n ,S 2

n =a 2n +1-λS n +1, ∴S 2n =(S n +1-S n )2-λS n +1,

∴S n +1(S n +1-2S n -λ)=0. ∵a n >0,∴S n +1>0,

∴S n +1-2S n -λ=0,∴S n +1=2S n +λ. (2)∵S n +1=2S n +λ, ∴S n =2S n -1+λ(n ≥2),

两式相减,得a n +1=2a n (n ≥2). ∵S 2=2S 1+λ,即a 2+a 1=2a 1+λ, ∴a 2=1+λ,由a 2>0,得λ>-1.

若{a n }是等比数列,则a 1a 3=a 2

2,

即2(λ+1)=(λ+1)2,得λ=1. 经检验,λ=1符合题意.

故存在λ=1,使得数列{a n }为等比数列.

12.(2020·合肥质检)已知等差数列{a n }的前n 项和为S n ,a 2=1,S 7=14,数列{b n }满足b 1·b 2·b 3·…·b n =2

n 2+n

2

.

(1)求数列{a n }和{b n }的通项公式;

(2)若数列{c n }满足c n =b n cos(a n π),求数列{c n }的前2n 项和T 2n . [解] (1)设等差数列{a n }的公差为d ,由a 2=1,S 7=14,

得??

?

a 1+d =1,7a 1+21d =14.

解得a 1=12,d =12,∴a n =n

2.

∵b 1·b 2·b 3·…·b n =2 n 2+n 2 =2 n (n +1)2

, ∴b 1·b 2·b 3·…·b n -1=2

n (n +1)

2

(n ≥2),

两式相除得b n =2n (n ≥2). 又当n =1时,b 1=2也适合上式,

∴b n =2n .

(2)∵c n =b n cos(a n π)=2n

cos ? ??

??

n 2π,

∴T 2n =2cos π2+22cosπ+23

cos 3π2+24cos(2π)+ (22)

1

cos (2n -1)π

2

+22n

cos(n π) =22cosπ+24cos(2π)+26cos(3π)+…+22n cos(n π) =-22+24-26+…+(-1)n ·22n =-4[1-(-4)n ]

1+4

=-4+(-4)n +1

5

.

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课标版云南省2019年中考英语总复习第二部分语法专题研究专题十三简单句试题(含解析)18

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