s3/S6=1/3

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等比数列S6/S3=(1-q^3)S3/S3S6/S3=(1-q^3)S3/S3为什么S6=(1-q

等比数列S6/S3=(1-q^3)S3/S3S6/S3=(1-q^3)S3/S3为什么S6=(1-q^3)S3同时乘以S3啊把分母去掉

已知等差数列{an} S3:S6=1/3S6:S12=?

已知等差数列{an}S3:S6=1/3S6:S12=?S3/S6=[(a1+a3)*3/2]/[(a1+a6)*6/2]=1/3(9/2)(a1+a3)=3(a1+a6)3(a1+a1+2d)=2(a1+a1+5d)6a1+6d=4a1+1

等比数列 S6/S3=3 则S9/S6=?

等比数列S6/S3=3则S9/S6=?由S6/S3=3得{1-q^6}/{1-q^3}=3所以q^3=2而S9/S6={1-q^9}/{1-q^6}=1-8/1-4=7/3这道题可根据一公式来解:即(S6-S3)^2=S3*(S9-S6),

设Sn是等差数列an的前n项和,若S3/S6=1/3,则S6/S7=?

设Sn是等差数列an的前n项和,若S3/S6=1/3,则S6/S7=?S3/S6=1/3S6=3S36a1+15d=3(3a1+3d)-3a1=-6da1=2d所以an=2d+(n-1)*d=(n+1)dSn=n(2+n+1)d/2=n(3

Sn是等差数列an前n项和,若S3/S6=1/3,则S6/S12=?

Sn是等差数列an前n项和,若S3/S6=1/3,则S6/S12=?Sn是等差数列an前n项和则S3,S6-S3,S9-S6,S12-S9也成等差数列因为S3/S6=1/3故S6=3S3所以S6-S3=2S3所以公差是d=(S6-S3)-S

设Sn是等差数列(An)的前n项和,若S3/S6=1/3,求S6/S12=?

设Sn是等差数列(An)的前n项和,若S3/S6=1/3,求S6/S12=?S3=(a1+a3)*3/2=(2a1+2d)*3/2=3a1+3dS6=(a1+a6)*6/2=(2a1+5d)*3=6a1+15dS3/S6=1/36a1+15

设Sn是等差数列{An}的前n项和,若S3/S6=1/3,则S6/S12=?

设Sn是等差数列{An}的前n项和,若S3/S6=1/3,则S6/S12=?S3/S6=1/3(3a1+3d)/(6a1+15d)=1/3(a1+d)/(2a1+5d)=1/3a1=2dS6/S12=(6a1+15d)/(12a1+66d)

设Sn是等差数列的前n项和,S3:S6=1:3,则S6:S9=

设Sn是等差数列的前n项和,S3:S6=1:3,则S6:S9=答案1:2方法一:因为Sn是等差数列的前n项和所以S3、S6-S3、S9-S6、……S(3n+3)-S(3n)成为等差数列设该数列为bn已知S3:S6=1:3,即3S3=S6S6

设Sn是等差数列{an}的前n项和,若S3/S6=1/3,则S6/S12=?

设Sn是等差数列{an}的前n项和,若S3/S6=1/3,则S6/S12=?3/10根据S3/S6=1/3,求出a1与d的比值,在代入S6/S12约去d即可

关于等差数列的一题已知在等差数列中,S3/S6=1/3,求S6/S12?

关于等差数列的一题已知在等差数列中,S3/S6=1/3,求S6/S12?设等差数列首项为a,公差为d:S3=3a+3dS6=6a+15dS12=12a+66d由题意:S3/S6=1/3即:(3a+3d)/(6a+15d)=1/3解得:a=2

设Sn是等差数列An的前n项和,若S3/S6=1/3,则S6/S12等于

设Sn是等差数列An的前n项和,若S3/S6=1/3,则S6/S12等于等差数列S3,S6-S3,S9-S6,S12-S9也成等差数列S3/S6=1/3,S6=3S3,S6-S3=2S3S9-S6=3S3,S9=6S3S12-S9=4S3,

设Sn是等差数列{an}的前n项和,若S3/S6=1/3,求S6/S12

设Sn是等差数列{an}的前n项和,若S3/S6=1/3,求S6/S12首先你要知道等差数列的顺次n项和性质即Sn,S2n-Sn,S3n-S2n成公差为n²d的等差数列则S6-S3=S3+9d由S3/S6得S3=9d则S6=27d

sn是等差数列{an}前n项和,若s3/s6=1/3,求s6/s12

sn是等差数列{an}前n项和,若s3/s6=1/3,求s6/s12设等差数列首项为a,公差为d,可以计算出S3=3a+3dS6=6a+15dS12=12a+66dS3/S6=1/3则(3a+3d)/(6a+15d)=1/3解得a=2d所以

1.设Sn是等差数列【An]的前n项和,若S3/S6=1/3,求S6/S12

1.设Sn是等差数列【An]的前n项和,若S3/S6=1/3,求S6/S12设等差数列的首项为a1,公差为d.则有(a1+2d)/(a1+5d)=1/3,解得d=-2a1那么S6/S12=(a1=5d)/(a1+11d)=(9a1)/(21

等比数列 a1=1 s6=4s3 求a4sudu

等比数列a1=1s6=4s3求a4sudus6/s3=(1-q^6)/(1-q^3)=1+q^3=4q^3=3a4=a1*q^3=3

等比数列 已知s6/s3=3/2 求s12/s9

等比数列已知s6/s3=3/2求s12/s9s6/s3=3/2s6-s3=(s3)/2S3,S6-S3,S9-S6,S12-S9是等比数列.公比(s6-s3)/s3=1/2.故S9-S6=(S3)/4,S12-S9=(S3)/8所以S9=7

等比数列前n项和Sn,S6/S3=3,S9/S6=?

等比数列前n项和Sn,S6/S3=3,S9/S6=?S3=a1(1-q^3)/(1-q)S6=a1(1-q^6)/(1-q)S9=a1(1-q^9)/(1-q)S6/S3=(1-q^6)/(1-q^3)=1+q^3=3q^3=2所以S9/S

等比数列的前项和Sn,若S6/S3=3,则S9/S6=?

等比数列的前项和Sn,若S6/S3=3,则S9/S6=?因为S6/S3=(1-q^6)/(1-q^3)=3(1-q^6)=3(1-q^3)令q^3=tt^2-3t+2=0t=1or2所以1.q^3=1s9/s6=9/6=3/22.q^3=2

在等比数列{an}中,若S6/S3=3,求S9/S6

在等比数列{an}中,若S6/S3=3,求S9/S6S6/S3=[a1(q^6-1)/(q-1)]/[a1(q^3-1)/(q-1)]=q^3+1=3S9/S6=[a1(q^9-1)/(q-1)]/[a1(q^6-1)/(q-1)]=(q^

等差数列(an)的前n项和sn,若s6:s3=3,求s9:s6

等差数列(an)的前n项和sn,若s6:s3=3,求s9:s6因为S6:S3=3所以S6-S3=2*S3又应为S3,S6-S3,S9-S6,成等差数列这个等差数列就是S32*S33*S3,所以S9-S6=3*S3所以S9=S6+S3=6*S