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2019-2020学年高中数学 第一章 数列 1.3 等比数列 1.3.1.1习题精选 北师大版必修5

2019-2020 学年高中数学 第一章 数列 1.3 等比数列 1.3.1.1 习题 精选 北师大版必修 5
1.若{an}是等比数列,则下列数列不是等比数列的是 ( )

A.{an+1} 答案:A

B.

C.{4an}

D.{

}

2.在等比数列{an}中,2a4=a6-a5,则公比是( A.0 C.-1 或 2 B.1 或 2 D.-1 或-2
3 5

)

解析:设公比为 q(q≠0),由已知得 2a1q =a1q -a1q ,

4

∴2=q2-q,∴q2-q-2=0, ∴q=-1 或 q=2.
答案:C 3.若一个等比数列的首项为 ,末项为 ,公比为 ,则这个数列的项数为( A.3 B.4 C.5 D.6 )

解析:在等比数列中,



,

∴n-3=1,即 n=4,故选 B.
答案:B 4.若数列{an}满足 an+1=4an+6(n∈N+)且 a1>0,则下列数列是等比数列的是( A.{an+6} C.{an+3} B.{an+1} D.{an+2} )

解析:由 an+1=4an+6 可得 an+1+2=4an+8=4(an+2),因为 a1>0, 所以 an>0,从而 an+2>0(n∈N+),因此 答案:D 5.在等比数列{an}中,若 a5·a6·a7=3,a6·a7·a8=24,则 a7·a8·a9 的值等于( A.48 B.72 C.144 D.192
3

=4,故{an+2}是等比数列.

)

解析:设公比为 q,由 a6·a7·a8=a5·a6·a7·q , 得 q = =8.
3

所以 a7·a8·a9=a6·a7·a8·q =24×8=192. 答案:D 6.数列{an}是公差不为 0 的等差数列,且 a1,a3,a7 为等比数列{bn}的连续三项,则数列{bn}的公 比为( A. ) B.4 C.2 D.

3

解析:∵a1,a3,a7 为等比数列{bn}中的连续三项,



=a1·a7.

设{an}的公差为 d,则 d≠0,

∴(a1+2d)2=a1(a1+6d),∴a1=2d. ∴公比 q=
答案:C 7.(2017 全国 3 高考)设等比数列{an}满足 a1+a2=-1,a1-a3=-3,则 a4= 解析:设{an}的公比为 q,则由题意,得 答案:-8 8.设数列{an}是等比数列,公比 q=2,则 解析:∵q=2,∴2a1=a2,2a3=a4, 的值是 解得
3

=2,故选 C.

.

故 a4=a1q =-8.

.



.

答案: 9.已知数列{an}满足 a9=1,an+1=2an(n∈N+),则 a5= 解析:由 an+1=2an(n∈N+)知,数列{an}是公比 q=

. =2 的等比数列.

所以 a5=a1q =

4

.

答案:

10.若数列{an}为等差数列,且 a2=3,a5=9,则数列 或“等比”). 解析:设{an}的公差为 d,则

一定是

数列(填“等差”

解得

于是 an=2n-1,

从而

=2·

,

设 bn=2·

,则

,

故 答案:等比 11. 解析:∵a1a9=

一定是等比数列.

导学号 33194017 在等比数列{an}中,a1·a9=256,a4+a6=40,则公比 q= .

q8,a4a6=a1q3·a1q5=

q8,

∴a1a9=a4a6.
可得方程组

解得

∴q2=

或 q = =4.

2

∴q=± 或 q=±2.

答案:-2,2,12.在等比数列{an}中,已知 a1=2,a4=16. (1)求数列{an}的通项公式; (2)若 a3,a5 分别为等差数列{bn}的第 3 项和第 5 项,试求数列{bn}的通项公式. 解(1)设{an}的公比为 q(q≠0),由已知得 16=2·q ,解得 q=2,∴an=a1·q =2×2 =2 .
3

n-1

n-1

n

(2)由(1)得 a3=8,a5=32,则 b3=8,b5=32, 设{bn}的公差为 d, 则有

解得

∴bn=-16+12(n-1)=12n-28.
13. 导学号 33194018 已知关于 x 的二次方程 anx -an+1x+1=0(n∈N+)的两根 α ,β
2

满足 6α -2α β +6β =3,且 a1=1. (1)试用 an 表示 an+1; (2)求证:数列 为等比数列;

(3)求数列{an}的通项公式. (1)解因为 α ,β 是方程 anx -an+1x+1=0(n∈N+)的两根,
2

所以

又因为 6α -2α β +6β =3,所以 6an+1-3an-2=0. 所以 an+1= an+ .

(2)证明因为 an+1= an+ ? an+1-

an-

为常数,且 a1-

,

所以

为等比数列.

(3)解令 bn=an- ,则{bn}为等比数列,公比为 ,首项 b1=a1-

,

所以 bn=

.

所以 an=bn+

.

所以数列{an}的通项公式为 an=

.

14.

导学号 33194019 容积为 a L(a>1)的容器盛满酒精后倒出 1 L,然后加满水,

再倒出 1 L 混合溶液后又用水加满,如此继续下去,问第 n 次操作后溶液的浓度是多少?当 a=2 时,至少应倒出几次后才可能使酒精浓度低于 10%? 解开始的浓度为 1,操作一次后溶液的浓度是 a1=1- .

设操作 n 次后溶液的浓度是 an,则操作 n+1 次后溶液的浓度是 an+1=an

.

所以{an}构成以 a1=1- 为首项,q=1- 为公比的等比数列.

所以 an=

,即第 n 次操作后溶液的浓度是

.

当 a=2 时,由 an=

,得 n≥4.

因此,至少应倒 4 次后才可以使酒精浓度低于 10%.


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