创新设计江苏专用理科高考数学二轮专题复习 填空题补偿练6
北京理工大学分数线-风流嫂嫂
补偿练6 数 列
(建议用时:40分钟)
1.设S
n
是
等差数列{a
n
}的前n项和,若S
7
=35,则a
4
等于
________.
7(a
1
+a
7
)
7×2a
4
解析
由题意,
=
2
=35,所以a
4
=5.
2
答案
5
2.在等比数列{a
n
}中,若a
4
,a
8
是
方程x
2
-3x+2=0的两根,则a
6
的值是________.
a
4
+a
8
=3>0,
解析 依题意
得
因此a
4
>0,a
8
>0,a
6
=a
4
a
8
=2.
a
4
a
8
=2>0,
答案 2 3.等差数列{a
n
}中,若a
1
+a
2
=2,a5
+a
6
=4,则a
9
+a
10
=_____
___.
解析 根据等差数列的性质,a
5
-a
1
=a
9
-a
5
=4d,a
6
-a
2
=a
10<
br>-a
6
=4d,∴(a
5
+a
6
)-(a
1
+a
2
)=8d,而a
1
+a
2
=2,a
5
+a
6
=4,∴8d=2,a
9
+a
10
=a<
br>5
+a
6
+
8d=4+2=6.
答案 6
4.
已知等比数列{a
n
}的前三项依次为a-1,a+1,a+4,则a
n
=_
_______.
解析 由题意得(a+1)
2
=(a-1)(a+4),解得a
=5,故a
1
=4,a
2
=6,所以a
n
6
n-1
3
n-1
4
2
=4·=4·
.
3
n-1
2
答案 4·
5.等差数列{a
n
}的前n项和为S
n
,且a
3<
br>+a
8
=13,S
7
=35,则a
8
=______
__.
2a
1
+9d=13,
解析
设a
n
=a
1
+(n-1)d,依题意
7a
1
+21d=35,
a
1
=2,
解得
所以a
8
=9.
d=1,
答案 9
6.已知等比
数列{a
n
}的公比为正数,且a
3
a
9
=2a
2
5
,a
2
=2,则a
1
=________.
2
解析 因为等比数列{a
n
}的公比为正数,且a
3
a<
br>9
=2a
5
,a
2
=2,所以由等比数
a
6
a
2
2
列的性质得a
2
6
=2a
5
,∴a
6
=2a
5
,公比q=
=2,a
1
==2.
aq
5
答案 2
16
S
8
7.设
S
n
是公差不为0的等差数列{a
n
}的前n项和,若a
1
=2a
8
-3a
4
,则
S
=
________.
8a
1
+28d
5S
8
解析 由已知得a
1
=2a
1
+14d-3a
1
-9d,∴a
1
=
2
d,又
S
=,将
16
16a
1
+120
d
5S
8
3
a
1
=
2
d代入化简得
S
=
10
.
16
3
答案
10
8.设数列{a
n
}是由正数组成的等比数列,S
n
为其前n项和,已知a
2
a
4
=1,S
3
=7,<
br>则S
5
=________.
解析 设此数列的公比为q(q>0),由已
知a
2
a
4
=1,得a
2
3
=1,所以a
3
=1.由
a
3
a
3
1
S
3
=7
,知a
3
+
q
+
q
2
=7,即6q
2-q-1=0,解得q=
2
,进而a
1
=4,所以S
5
1
5
4
1-
2
31
=
1
=
4
.
1-
2
31
答案
4
3
S
4
9.设等比数列{a
n
}的公比q=2,前n项的和为S
n,则
a
的值为________.
a
1
(1-q
4
)
S
4
15
解析
∵S
4
=
,a
3
=a
1
q
2
,∴
a
=
4
.
3
1-q
15
答案
4
10.已知各项不为0的等差数列{a
n
}满足a
4
-2a
2
7
+3a
8
=0,数列{b
n
}
是等比数列,
且b
7
=a
7
,则b
2
b
8
b
11
=________.
2
解析
设等差数列的公差为d,由a
4
-2a
2
7
+3a
8
=0,得a
7
-3d-2a
7
+3(a
7
+d)
=0,从而有a
7
=2或a
7
=0(a
7
=b
7<
br>,而{b
n
}是等比数列,故舍去),设{b
n
}的公
比为q
,则b
7
=a
7
=2,
b
7
∴b
2
b
8
b
11
=
q
5
·b
7
q·b
7
q
4
=(b
7
)
3<
br>=2
3
=8.
答案 8
1+2+3+…+n
1
,
则数列{}的前n项和为
n
a
n
a
n
+
1
11.已知数列{a
n
}满足a
n
=
__________.
1+2+3+…+nn+1
14
解析 a
n
=
=
,==
n2
a
n
a
n+1
(n+1)(n+2)
1
11
1
1111
--+-+…+-
,所求的前n项和为4
2334
=
4
<
br>n+1n+2n+1n+2
1
1
2n
-
=
4
2
.
n+2
n+2
答案
2n
n+2
12.设
等差数列{a
n
}的前n项和为S
n
,且a
1
>0,a3
+a
10
>0,a
6
a
7
<0,则满足S<
br>n
>0的最大自然数n的值为________.
解析 ∵a
1
>
0,a
6
a
7
<0,∴a
6
>0,a
7
<
0,等差数列的公差小于零,又a
3
+
a
10
=a
1
+a
12
>0,a
1
+a
13
=2a
7
<0,∴S
12
>0,S
13
<0,∴满足S
n
>0的最大
自
然数n的值为12.
答案 12
13.已知函数f(x)=(1-3m)x+
10(m为常数),若数列{a
n
}满足a
n
=f(n)(n∈N
*
),且
a
1
=2,则数列{a
n
}前100项的和为___
_____.
解析 ∵a
1
=f(1)=(1-3m)+10=2,∴m=3,∴
a
n
=f(n)=-8n+10,∴S
100
=-8(1+2
…+100)+10×100=-8×
答案 -39 400
101×100
+10×100=-39 400.
2
14.整数数列{a
n
}满足a
n
+
2
=a
n
+
1<
br>-a
n
(n∈N
*
),若此数列的前800项的和是2
013,
前813项的和是2 000,则其前2 014项的和为________.
解析 a
3
=a
2
-a
1
,a
4
=a3
-a
2
,a
5
=a
4
-a
3
,a
6
=a
5
-a
4
,a
7
=a
6
-a
5
,…,∴a
1
=a
7
,a
2<
br>=a
8
,a
3
=a
9
,a
4
=a<
br>10
,a
5
=a
11
,…,{a
n
}是以6
为周期的数列,且
有a
1
+a
2
+a
3
+a
4
+a
5
+a
6
=0,S
800
=a
1
+a
2
=2 013,S
813
=a
1
+a
2
+a
3
=2 000,
a
1
-a
2
=13,
a
3
=-13,∴
∴a
2<
br>=1 000,S
2 014
=a
1
+a
2
+a3
+a
4
=a
2
+a
3
=1
a
1
+a
2
=2
013,
000+(-13)=987.
答案 987