等比数列材料题目及答案

等比数列材料题目及答案考试时间：120分钟 总分：100分 年级/班级：初中二年级下学期

等比数列材料题目及答案

一、选择题

1.已知等比数列{a_n}中，a_1=2，a_3=8，则该数列的公比q等于（）

A.2

B.4

C.8

D.16

2.若一个等比数列的前三项依次为a，ar，ar^2，则该数列的前n项和S_n等于（）

A.a(1-r^n)/(1-r)

B.ar(1-r^n)/(1-r)

C.a(1+r^n)/(1+r)

D.ar(1+r^n)/(1+r)

3.已知等比数列{a_n}中，a_4=27，a_7=81，则a_5+a_6的值等于（）

A.36

B.45

C.54

D.63

4.若等比数列{a_n}的前n项和为S_n，公比为q，则当q≠1时，S_n的表达式为（）

A.a_1q^n-a_1

B.a_1(1-q^n)/(1-q)

C.a_1(1+q^n)/(1+q)

D.a_1q^n+a_1

5.已知等比数列{a_n}中，a_1=1，a_4=16，则该数列的通项公式a_n等于（）

A.2^(n-1)

B.2^(n+1)

C.4^(n-1)

D.4^(n+1)

6.若一个等比数列的前两项分别为3和6，则该数列的第三项等于（）

A.9

B.12

C.18

D.24

7.已知等比数列{a_n}中，a_2=6，a_5=162，则该数列的公比q等于（）

A.3

B.4

C.5

D.6

8.若等比数列{a_n}的前n项和为S_n，公比为q，则当q=1时，S_n的表达式为（）

A.na_1

B.n(a_1+a_n)/2

C.a_1q^n-a_1

D.a_1(1-q^n)/(1-q)

9.已知等比数列{a_n}中，a_1=5，a_3=10，则该数列的公比q等于（）

A.1

B.2

C.5

D.10

10.若一个等比数列的前三项依次为1，2，4，则该数列的第四项等于（）

A.8

B.16

C.32

D.64

二、填空题

1.已知等比数列{a_n}中，a_1=3，a_4=81，则该数列的公比q等于_______。

2.若一个等比数列的前两项分别为2和6，则该数列的第三项等于_______。

3.已知等比数列{a_n}中，a_2=4，a_5=64，则该数列的公比q等于_______。

4.若等比数列{a_n}的前n项和为S_n，公比为q，则当q≠1时，S_n的表达式为_______。

5.已知等比数列{a_n}中，a_1=1，a_4=16，则该数列的通项公式a_n等于_______。

6.若一个等比数列的前三项依次为a，ar，ar^2，则该数列的前n项和S_n等于_______。

7.已知等比数列{a_n}中，a_4=27，a_7=81，则a_5+a_6的值等于_______。

8.若等比数列{a_n}的前n项和为S_n，公比为q，则当q=1时，S_n的表达式为_______。

9.已知等比数列{a_n}中，a_1=5，a_3=10，则该数列的公比q等于_______。

10.若一个等比数列的前三项依次为1，2，4，则该数列的第四项等于_______。

三、多选题

1.下列哪个选项是等比数列的性质？（）

A.从第二项起，每一项与它的前一项的比等于同一个常数

B.从第一项起，每一项与它的后一项的比等于同一个常数

C.若数列为等比数列，则其任意连续三项的积也成等比数列

D.若数列为等比数列，则其任意连续三项的平方也成等比数列

2.关于等比数列的前n项和S_n，下列哪个选项是正确的？（）

A.当q=1时，S_n=na_1

B.当q≠1时，S_n=a_1(1-q^n)/(1-q)

C.当q=-1时，S_n=0

D.当q=1时，S_n=a_1q^n-a_1

3.已知等比数列{a_n}中，a_1=2，a_3=8，则下列哪个选项是正确的？（）

A.该数列的公比q等于2

B.该数列的公比q等于4

C.该数列的通项公式a_n等于2^n

D.该数列的通项公式a_n等于2^(n-1)

4.关于等比数列的通项公式a_n，下列哪个选项是正确的？（）

A.a_n=a_1q^(n-1)

B.a_n=a_1q^n

C.a_n=a_1+(n-1)d

D.a_n=a_1+(n-1)q

5.已知等比数列{a_n}中，a_4=27，a_7=81，则下列哪个选项是正确的？（）

A.该数列的公比q等于3

B.该数列的公比q等于9

C.a_5+a_6的值等于54

D.a_5+a_6的值等于72

四、判断题

1.等比数列的任意一项都不能为0。

2.若等比数列{a_n}的前n项和为S_n，公比为q，则当q=1时，S_n=na_1。

3.等比数列的通项公式a_n=a_1q^(n-1)中，n必须为正整数。

4.若一个数列为等比数列，则其任意连续三项的积也成等比数列。

5.等比数列的前n项和S_n公式中，当q≠1时，S_n=a_1(1-q^n)/(1-q)。

6.若等比数列{a_n}中，a_1=2，a_3=8，则该数列的公比q等于2。

7.等比数列的任意一项a_n可以表示为a_n=a_mq^(n-m)。

8.若等比数列{a_n}的前n项和为S_n，公比为q，则当q=-1时，S_n=0。

9.等比数列的公比q必须为正数。

10.若一个数列为等比数列，则其任意连续三项的平方也成等比数列。

五、问答题

1.已知等比数列{a_n}中，a_1=3，a_4=24，求该数列的公比q和通项公式a_n。

2.若一个等比数列的前两项分别为1和2，求该数列的前五项和S_5。

3.已知等比数列{a_n}中，a_2=6，a_5=162，求该数列的公比q和前五项和S_5。

试卷答案

一、选择题答案及解析

1.B解析：根据等比数列的定义，a_3=a_1*q^2。将a_1=2，a_3=8代入，得8=2*q^2，解得q^2=4，所以q=2或q=-2。由于选项中没有负数，故q=2。

2.B解析：等比数列的前n项和S_n公式为S_n=a_1*(1-q^n)/(1-q)（当q≠1时）。根据题意，前三项依次为a，ar，ar^2，所以a_1=a，q=r。代入公式得S_n=ar*(1-r^n)/(1-r)。

3.A解析：根据等比数列的定义，a_7=a_4*q^3。将a_4=27，a_7=81代入，得81=27*q^3，解得q^3=3，所以q=3^(1/3)=3^(1/3)。a_5+a_6=a_4*q+a_4*q^2=27*3^(1/3)+27*(3^(1/3))^2=27*3^(1/3)+27*3^(2/3)=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^(1/3))=27*3^(1/3)*(1+3^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