1.若数列{an}的前n项和Sn=log(1/10)(1+n),则a10+a11+...+a99=.2.在等差数列中,a(2n)/an=(4n-1)/(2n-1),则S(2n)/Sn=.3.数列{an}中,a1=-20,a(n+1)=an+4,则|a1|+|a2|+...+|a20|=.4.设数列{an}是等差数列,且a2=-6,a8=6,前n项和为S

Question

1.若数列{an}的前n项和Sn=log(1/10)(1+n),则a10+a11+...+a99=___.
2.在等差数列中,a(2n)/an=(4n-1)/(2n-1),则S(2n)/Sn=___.
3.数列{an}中,a1=-20,a(n+1)=an+4,则|a1|+|a2|+...+|a20|=___.
4.设数列{an}是等差数列,且a2=-6,a8=6,前n项和为Sn,当Sn最大时,则n=___.
5.在等差数列{an}中,若a10=10,a19=100,Sn=0,则n=___.
6.已知等差数列{an},an=2n-1,则S1+S2+S3+S4+S5=___.
7.在等差数列{an}中,d不等于0,已知S100=100S10,则a100/a10=___.
8.等差数列{an}中,d=1/2,且S100=145,则a1+a3+a5+...+a99=___.
9.在等差数列{an}中,S10=100,S100=10,则S110=___.
10.在等差数列{an}中,a3=12,S14>0,S15

Answer 1

1.若数列{an}的前n项和Sn=log(1/10)(1+n),则a10+a11+...+a99=___.
a10+a11+...+a99=S99-S9
=log(1/10)(1+99)-log(1/10)(1+9)
=log(1/10)10
=-1
2.在等差数列中,a(2n)/an=(4n-1)/(2n-1),则S(2n)/Sn=___.
因为an为等差数列,
/(2n-1)
所以a1+(2n-1)d/a1+(n-1)d=(4n-1)/(2n-1)
2a1=d
所以an=a1+(n-1)d=（2n-1)a1
an=a1+(2n-1)d=（4n-1)a1
S(2n)/Sn
=[（a1+a2n)*2n/2]/[(a1+an)*n/2]
=（4n*2n/2)/（2n*n/2)
=4
3.数列{an}中,a1=-20,a(n+1)=an+4,则|a1|+|a2|+...+|a20|=___.
a(n+1)=an+4,an是以4为公差的等差数列.
an=4n-24
显然,当n>6时an>0
所以|a1|+|a2|+...+|a20|=S20-2S6
=(-20+56)*20/2+(20+0)/2*2
=380
4.设数列{an}是等差数列,且a2=-6,a8=6,前n项和为Sn,当Sn最大时,则n=___.
数列{an}是等差数列.所以可设an=a1+(n-1)d
-6=a1+d
6=a1+7d
解得a1=-8,d=2
所以an=2n-10
Sn为增函数无最大值.只能求的最小值,显然当全为负数时最小,则
an=2n-10<=0
所以前4,5项的值最小
5.在等差数列{an}中,若a10=10,a19=100,Sn=0,则n=___.
数列{an}是等差数列.所以可设an=a1+(n-1)d
10=a1+9d
100=a1+18d
解得a1=-80,d=10
所以an=10n-90
Sn=（-80+10n-90）*n/2=（10n-170）n/2=0
所以n=17
6.已知等差数列{an},an=2n-1,则S1+S2+S3+S4+S5=___.
an=2n-1.则sn=(1+2n-1)n/2=n^2
则S1+S2+S3+S4+S5=1^2+2^2+...+5^2=55
7.在等差数列{an}中,d不等于0,已知S100=100S10,则a100/a10=___.
数列{an}是等差数列.所以可设an=a1+(n-1)d
S100=100S10
(2a1+99d)*100/2=100*(2a1+9d)*10/2
解得a1=109/2d
则a100/a10=(a1+99d)/(a1+9d)=307/127
8.等差数列{an}中,d=1/2,且S100=145,则a1+a3+a5+...+a99=___.
a1,a3,a5,..+a99成公差为1的等差数列.
S100=a1+a2+...+a100
S1=a1+a3+a5+...+a99
S2=a2+a4+a6+...+a100
S2-S1=(a2-a1)+(a4-a3)+...+(a100-a99)=50/2=25
S1+S2=S100=145
俩式相减得
2S1=120
所以S1=60
9.在等差数列{an}中,S10=100,S100=10,则S110=___.
10.在等差数列{an}中,a3=12,S14>0,S15<0,则n=___时Sn有最大值.