Lisa's birthday is on October 18th. It's on Wednesday. She is going to have a birthday party. She wants to invite Peter, Gao Wei and Li Yan to her birthday party. They can celebrate after school. Today Lisa's parents go shopping with Lisa. They want to buy a My dream is to work in a hospital.I thin that a(n) __1__ is one of the most important jobs in the world.__2__ people are busy working every day 第一讲 等差数列、等比数列
限时规范训练
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一、选择题
1．已知数列{an}，若点(n，an)(n∈N)在经过点(8,4)的定直线l上，则数列{an}的前15项和S15＝
B．32D．120
( )A．12C．60
解析：∵点(n，an)在定直线上，∴数列{an}是等差数列，且a8＝4，
∴S15＝
＋2
2a8×15
＝＝15a8＝60.
2
答案：C
2．已知各项不为0的等差数列{an}满足a4－2a27＋3a8＝0，数列{bn}是等比数列，且b7＝a7，则b2b8b11
B．2D．8
等于( )
A．1 C．4
解析：∵a4－2a27＋3a8＝0，∴2a27＝a4＋3a8，∴2a27＝a5＋a7＋2a8＝a5＋a7＋a7＋a9，即2a27＝4a7，
∴a7＝2，∴b7＝2，又∵b2b8b11＝b6b8b7＝b27b7＝(b7)＝8，故选D.
答案：D
3
3．在等差数列{an}中，an>0，且a1＋a2＋…＋a10＝30，则a5·a6的最大值等于()
B．6D．36
A．3C．9
30?a5＋a6?2＝?6?2＝9.
解析：∵a1＋a2＋…＋a10＝30，得a5＋a6＝＝6，又an>0，∴a5·a6≤????5?2??2?
答案：C
4．设等差数列{an}满足a2＝7，a4＝3，Sn是数列{an}的前n项和，则使得Sn＞0的最大的自然数n
B．10D．12
是( )
A．9C．11
3－7解析：∵{an}的公差d＝＝－2，∴{an}的通项为an＝7－2(n－2)＝－2n＋11，∴{an}是递减数
4－2
列，
a5＋a6
且a5＞0＞a6，a5＋a6＝0，于是S9＝9a5＞0，S10＝·10＝0，S11＝11a6＜0，故选A.
2
答案：A
5．在等比数列{an}中，a1＋an＝34，a2·an－1＝64，且前n项和Sn＝62，则项数n等于( )
Lisa's birthday is on October 18th. It's on Wednesday. She is going to have a birthday party. She wants to invite Peter, Gao Wei and Li Yan to her birthday party. They can celebrate after school. Today Lisa's parents go shopping with Lisa. They want to buy a My dream is to work in a hospital.I thin that a(n) __1__ is one of the most important jobs in the world.__2__ people are busy working every day
B．5D．7
A．4 C．6
解析：设等比数列{an}的公比为q，由a2an－1＝a1an＝64，又a1＋an＝34，解得a1＝2，an＝32或a1＝32，an＝2.当a1＝2，an＝32时，Sn＝
－
1－q
＝a1－anq2－32qn＝＝62，解得q＝2.又an＝a1q1－q1－q
由an＝a1q所以2×2
n－1
n－1
－1
，
1n＝2＝32，解得n＝5.同理，当a1＝32，an＝2时，由Sn＝62，解得q＝.
2
?1?n－1?1?n－11?1?4
＝32×??＝2，得??＝＝??，即n－1＝4，n＝5.综上，项数n等于5，故选
16?2??2??2?
B.答案：B
S12S10
6．在等差数列{an}中，a1＝－2015，其前n项和为Sn，若－＝2，则S2016的值等于( )
1210
B．2 015
D．0
A．－2 015 C．2 016
12×1110×9
解析：设数列{an}的公差为d，S12＝12a1＋d，S10＝10a1＋d，
2212×11
12a1＋d2S1211S109S12S10
所以＝＝a1＋d. ＝a1＋d，所以－＝d＝2，
121221021210
2 015×2 016
所以S2 016＝2 016×a1＋d＝0.
2
答案：D
S1S2S15
7．设等差数列{an}的前n项和为Sn，且满足S17＞0，S18＜0，则，，…，中最大的项为( )
a1a2a15
解析：因为{an}是等差数列，所以S17＝
B.D.1＋
2
S8a8
A.C.
＝17a9＞0，a9＞0，S18＝＋
2
S9a9
S7 a7S9 a9
S10a10
＝9(a9
＋a10)＜0，
a10＜0，即该等差数列前9项均是正数项，从第10项开始是负数项，则最大，故选C.
答案：C
8．正项等比数列{an}中，a2＝8,16a24＝a1a5，则数列{an}的前n项积Tn中的最大值为( )
B．T4
A．T3
Lisa's birthday is on October 18th. It's on Wednesday. She is going to have a birthday party. She wants to invite Peter, Gao Wei and Li Yan to her birthday party. They can celebrate after school. Today Lisa's parents go shopping with Lisa. They want to buy a My dream is to work in a hospital.I thin that a(n) __1__ is one of the most important jobs in the world.__2__ people are busy working every day D．T6C．T5
12a41
解析：设正项等比数列{an}的公比为q(q＞0)，则16a24＝a1a5＝a2a4＝8a4，a4＝，q＝＝，又
2a216
q＞0，则q＝，an＝a2qn－2＝8×??n－2＝27－2n，则Tn＝a1a2…an＝25＋3＋…＋(7－2n)＝2n(6－n)，当n＝3时，
4
14
?1???
n(6－n)取得最大值9，此时Tn最大，即(Tn)max＝T3，故选A.
答案：A
二、填空题
a2＝b2
9．(2017·高考北京卷)若等差数列{an}和等比数列{bn}满足a1＝b1＝－1，a4＝b4＝8，则
________.
解析：设等差数列{an}的公差为d，等比数列{bn}的公比为q，
a4－a18－－
则由a4＝a1＋3d，得d＝＝33
＝3，
b4833
由b4＝b1q得q＝＝＝－8，∴q＝－2.
b1－1
∴
a2a1＋d－1＋3＝＝b2b1q－－
＝1.
答案：1
10．若等比数列{an}的各项均为正数，且a10a11＋a9a12＝2e，则ln a1＋ln a2＋…＋ln a20＝________.
解析：因为a10a11＋a9a12＝2a10a11＝2e，所以a10a11＝e.
10
5
5
5
所以ln a1＋ln a2＋…＋ln a20＝ln(a1a2…a20)＝ln[(a1a20)·(a2a19)·…·(a10a11)]＝ln(a10a11)
5
＝10ln(a10a11)＝10ln e＝50ln e＝50.
答案：50
11．等比数列{an}的首项为2，公比为3，前n项和为Sn.若log3?
?1?2
＋
?＝9，则1＋4取最?nm?
小值时，S2＝________.＋
解析：由题意可得an＝2×3
n－1
，Sn＝
－
1－3
＝3－1，所以log3?
n?1
?2
?＝log3n＋4m－1＝?3?
14?14??n2m?172n2m172255
n＋4m－1＝9，所以n＋4m＝10，所以＋＝?＋??＋?＝＋＋≥＋2×＝＝，当
nm?nm??105?105m5n105102
且仅当m＝n时取等号，∴n＝2，∴a2＝2×3＝6，∴S2＝2＋6＝8.
答案：8
n1
12．(2017·高考全国卷Ⅱ)等差数列{an}的前n项和为Sn，a3＝3，S4＝10，则? ＝________.
Skk＝1
2018年高考数学二轮复习第一部分专题三数列第一讲等差数列等比数列习题
