2013届高考人教A版文科数学一轮强化训练5.2等差数列及其前n项和

1.等差数列{n a }的前n 项为为n S ,且3164S a =,=,则公差d 等于( ) A.1 B.53
C.2
D.3
答案:C
解析:∵31336()2
S a a ==+且31124a a d a =+,=,
∴d=2.故选C.
2.已知{n a }为等差数列,且743210a a a -=-,=,则公差d 等于( ) A.-2 B.12
-
C.12
D.2
答案:B
解析:7433242()21a a a d a d d -=+-+==-,解得12
d =-.
3.如果等差数列{n a }中34512a a a ,++=,那么1a +2a +7a 等于( )
A.14
B.21
C.28
D.35
答案:C
解析:345443124a a a a a ++==,=,
∴12a a ++…747()
177282
a a a a ++===. 4.设等差数列{n a }的前n 项和为n S ,若972S =,则2a +49a a += . 答案:24
解析:∵{n a }是等差数列,由972S =,得959S a =,5a =8, ∴24929456()()a a a a a a a a ++=++=++
4a =5324a =.
5.在等差数列{n a }中35276a a a ,=,=+,则6a = . 答案:13
解析:设等差数列{n a }的公差为d,则由已知得 1112746a d a d a d +=,
⎧⎨
+=++.⎩ 解得
132a d =,
⎧⎨
=.
⎩ 所以61a a =
+
6.已知曲线C:xy-4x+4=0,数列{n a }的首项14a =,且当
2n ≥时,点1()n n a a -,恒在曲线C 上,
且n b =
12a
n
,-试判断数列{n
b }是否是等差数列?并说明理由.
解:∵当2n ≥时,点1()n n a a -,恒在曲线C 上, ∴11440n n n a a a ---+=. 由12n b a
n
=
-得: 当2n ≥时111122422111a a n n b b n n a a a a a a
n n n n n n --,-=-=
-----+--- 11142244222
111
a a a a
n n n n a a a a a n n n n n ----==
=---+--+---. ∴数列{n b }是公差为12
-的等差数列.
题组一 等差数列的基本运算
1.在等差数列{n a }中,已知12411039n a a a a =,+=,=,则n 等于( ) A.19 B.20 C.21 D .22
答案:B
解析:∵2411310a a a d a d +=+++=,∴d=2. 由1(1)n a a n d =+-=39,解得n=20.
2.等差数列{n a }的公差不为零,首项121a a =,是1a 和5a 的等比中项,则数列的前10项和是( ) A.90 B.100
C.145
D.190
答案:B
解析:设公差为d,则2
(1)1(14)d d +=⋅+. 因为0d ≠,解得d=2,∴10100S =.
3.等差数列{n a }的前n 项和为n S ,且53655S S -=,则4a = . 答案:13
解析:∵513151033S a d S a d =+,=+, ∴5311653060(1515)S S a d a d -=+-+=
115a +144515(3)155d a d a =+==,
∴413
a =.
4.已知数列{n a }是等差数列3410118a a a ,=,+=,则首项1a = . 答案:-3
解析:∵41033()(7)28a a a d a d d +=+++=+=18, ∴d=2.
∴1323a a d =-=-.
另解,∵7410218a a a =+=,∴79a =.
∴公差13
7391223734
a a
d a a d --===,=-=--. 题组二 等差数列性质的应用
5.等差数列{n a }的前n 项和为n S ,若2812a a +=,则9S 等于( ) A.54
B.45
C.36
D.27
答案:A
解析:99()9()19285422
a a a a S ++=
==. 6.已知等差数列{n a }的前n 项和为n S ,若4518a a =-,则8S 等于( ) A.68
B.72
C.54
D.90
答案:B
解析:∵4518a a =-,∴4518a a +=.
∴88()8()18457222
a a a a S ++=
==. 7.已知{n a }是等差数列67782028a a a a ,+=,+=,则该数列前13项和13S 等于( ) A.156
B.132
C.110
D.100
答案:A
解析:由67782028a a a a +=,+=知7448a =,∴712a =,故13713156S a ==,选A. 8.已知{n a }是等差数列451555a S ,=,=,则过点34(3)(4)P a Q a ,,,的直线的斜率为( ) A.4 B.14
C.-4
D.-14
答案:A
解析:∵{n a }是等差数列451555a S ,=,=, ∴153********a a a a +=,=,=.
∴43443
PQ
a a
k -==,-选A.
9.若等差数列{n a }的前5项和525S =,且23a =,则7a 等于( ) A.12 B.13 C.14 D.15
答案:B
解析:535S a =,∴35a =,∴d=2. ∴773213a =+⨯=,故选B.
10.设等差数列{n a }的前n 项和为n S ,若981S =,则2a +58a a += .
答案:27
解析:∵99()9()19288122
a a a a S ++=
==, ∴285218a a a +==. 即2585327a a a a ++==. 题组三 证明数列是等差数列
11.已知数列{n a }和{n b }满足1121(1)1n n n n n a a a a b a +=,-=-,=-.求数列{n b }的通项公式. 解:由1n n b a =-得1n n a b =+代入11(1)n n n a a a +-=-得1(1)n n n b b b +=+, 整理得11n n n n b b b b ++-=,
∵0(n b ≠否则1n a =,与12a =矛盾), 从而得
1
111b b n n
-=,+
∵1111b a =-=,
∴数列{1n
b }是首项为1,公差为1的等差数列.
∴1n b n
=,即1n b n
=.
12.已知{n a }是以a 为首项,q 为公比的等比数列n S ,为它的前n 项和. (1)当134S S S ,,成等差数列时,求q 的值;
(2)当m n l S S S ,,成等差数列时,求证:对任意自然数m k n k l k k a a a +++,,,也成等差数列.
解:(1)由已知1
n n a aq -,=,因此13(1S a S a =,=+
2
q q +234)(1)S a q q q ,=+++.
当134S S S ,,成等差数列时1432S S S ,+=,可得3
2
aq aq aq =+. 化简得2
10q q --=.
解得q =
(2)证明:若q=1,则{n a }的每项n a a =,此时m k a +、n k a +、l k a +显然成等差数列. 若1q ≠,由m n l S S S ,,成等差数列可得m S +l S =2n S ,
即(1)(1)2(1)111
m l n a q a q a q q q q ---+=
---. 整理得2m l n
q q q +=.
因此11
()22k m l n k m k l k n k a a aq q q aq
a -+-+++,+=+==. 所以m k a +,、n k a +、l k a +也成等差数列。