公差为3,a4 a6 a8=48,求通项公式
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an=3n-1a(n+1)=3(n+1)-1=3n+2公差d=a(n+1)-an=(3n+2)-(3n-1)=3
a1=3*1-4=-1d=3a1-d=-4
已知等差数列{an}的公差d=3且a1,a3,a4成等比数列∴a3²=a1×a4∴(a2+d)²=(a2-d)×(a2+2d)∴(a2+3)²=(a2-3)×(a2+6)
a1:a3=a3:a4a3=a1+3*2a4=a1+3*3a1:a1+3*2=a1+3*2:a1+3*336=-3a1a1=-12a2=a1+3=-12+3=-9
方法很多,我就说一个最容易理解的(当不一定是最简便的)根据sn,求出s1=4,s2=12,所以a1=s1=4,a1+a2=14,这样就可以把a2求出来=14-a1=10公差=a2-a1=6
由等差数列公式an=a1+(n-1)d可得an=nd+a1-d对照知d=-3a1-d=5所以a1=2
an=2-3na(n-1)=2-3(n-1)=2-3n+3=5-3nd=an-a(n-1)=2-3n-(5-3n)=2-3n-5+3n=-3所以公差d=-3
an=3+2(n-1)=2n+1lim[1/(a1a2)+1/(a2a3)+...+1/(a(n-1)an)]=lim(1/2)[1/3-1/5+1/5-/7+...+1/(2n-1)-1/(2n+1
因为a1+a5=a2+a4=4,所以:a2a4=3a2+a4=4解方程组:a2=1a4=3或者a2=3a4=1a4-a2=2d=2,或者a4-a2=-2d=1,或者d=-1
n+1-bn=3an+1+4b-(3an+4b)=3an+1-3an=3d所以是公差为3d的等差数列~
a1=-2d=-3
∵等差数列{an}的首项a1=3,公差d=2,∴前n项和Sn=na1+n(n−1)2d=3n+n(n−1)2×2=n2+2n(n∈N*),∴1Sn=1n2+2n=1n(n+2)=12(1n−1n+2)
(1)∵等差数列{an}中,a1=23,且a6=a1+5d>0,a7=a1+6d<0,∴23+5d>0,且23+6d<0,解得:-235<d<-236,又d为整数,∴d=-4;(2)∵a1=23,d=
(1)因为a4,a5,a8成等比数列,所以a52=a4a8.设数列{an}的公差为d,则(3+3d)2=(3+2d)(3+6d)化简整理得d2+2d=0.∵d≠0,∴d=-2.于是an=a2+(n-2
公差为3则a3=a1+2*3=a1+6a4=a1+3*3=a1+9a1,a3,a4成等比数列则(a3)^2=a1*a4(a1+6)^2=a1*(a1+9)a1^2+12a1+36=a1^2+9a1a1
a(2)=3-2×2=-1d=a2-a1=-2s(n)=[(a1+an)×n]/2=〔(1+3-2n)×n〕/2=2n-n×n另外验算:an=s(n)-s(n-1)=2n-n×n-2(n-1)+(n-
4设边从小到大为a,b,c则1/2absin120º=15√3得ab=60得b=60/a①a+c=2b得c=2b-a=120/a-a②﹙a²+b²-c²﹚/2a
a3+a5=2a4所以,原式化为:3a6=3a4+12即:a6-a4=4即:2d=4得:d=2
sn=n(a1+an)/2sn=a1+n(n-1)d/2解出a1=-7n=7a3=a1+2d=-1
an=11=a1+(n-1)d=a1+3(n-1)a1=14-3n代入下面sn=14=a1*n+n(n-1)d/2得(3n-4)(n-7)=0所以n=7a3=a7-(7-3)d=11-4*3=-1