在等差数列an中a不等于零
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/24 07:43:44
额..a1=da2=2dq=a2/a1=2d/d=2.
因为Sm/Sn=m^2/n^2,所以{[2a1+(m-1)d]*m}/{[2a1+(n-1)d]*n}=m^2/n^2,[2a1+(m-1)d]/[2a1+(n-1)d]=m/n,2a1n+(m-1)
1、设公差为d,则n=m+(m-n)d=>d=-1所以am+n=am+nd=02、x1-x2=(b-a)/3y3-y1=2(b-a)/4=(b-a)/2所以(y3-y1)/(x2-x1)=3/2
原问题即:有两个数列,{An}{Bn},若Sm:Sn=m^2:n^2求Am:Bn(公差分别为d1,d2)Sm=A1d+0.5*m(m-1)d1=0.5d1m^2+m(A1-0.5d1)Sn=0.5d2
给评价在答再问:别坑我呀,先给个好评了啊再问:你不会真是骗人的吧?亏我信你,现在人都这样儿么?再答:对不起我上课了
∵数列{an}是等差数列,∴a1=a3-2d=8-2da5=a3+2d=8+2da7=a3+4d=8+4d又∵a1,a5,a7成等比数列,∴a5²=a1·a7即﹙8+2d﹚²=﹙8
(1)a3=a1+2d,a7=a1+6d,所以a1*a7=a3*a3,即a1*(a1+6d)=(a1+2d)*(a1+2d)解得d=1(2)Sn=(1/2)n^2+(3/2)n,又a3=a1+2d=4
设An=A1+(n-1)d则A2=A1+dA4=A1+3d因为A2是A1与A4等比中项故(A2)²=A1A4即(A1+d)²=A1(A1+3d)d²=A1d因为d不为0,
设a=ab=a+dc=a+2d(d不等于0)用反证法证明设1/a,1/b,1/c是等差数列则2(1/b)=1/a+1/c2/b=2/(a+d)1/a+1/b=1/a+1/(a+2d)很明显不相等则假设
由题可知(b2)^2=b1xb3把a1=b1=1,a2=b2,a4=b3带入上式(a2)^2=a1xa4即(a1+d)^2=a1x(a1+3d)整理d=0舍掉或者d=a1=1则an=n所以b1=1b2
因为a5=a1+4d,a9=a1+8d,a15=a1+14d且a5a9a15成等比数列所以(a1+8d)^2=(a1+4d)(a1+14d)即(a1)^2+16a1*d+64d^2=(a1)^2+18
a8+a14=2a1+20d=0a1=-10d0Sn=na1+n(n-1)d/2=-10nd+n^2d/2-nd/2=(d/2)*n^2-(21d/2)n,对称轴是n=21/2=10.5所以,当n=1
假设1/a,1/b,1/c能成等差数列则2/b=1/a+1/c=(a+c)/ac.又a,b,c成等差数列,所以2b=a+c.带入上式,得2/b=2b/ac.即1/b=b/ac,所以b²=ac
∵a1=81,d=-7,∴an=81+(n-1)×(-7)=88-7n,由an=88-7n≥0,解得n≤1247,∴最接近零的是第13项,故选C.
a(n+1)-an^2+a(n-1)=0a(n-1)+a(n+1)=an^2由于是等差数列则a(n-1)+a(n+1)=2an所以an^2=2anan=2或an=0,不合题意,舍.所以an=2则S(2
a(n)=1+(n-1)da(n+1)=1+ndSn=(1+an)n/2=(2+nd-d)n/2(1+Sn)/(n(1-a(n+1)))=-((4+nd-d)/n)/(2n(nd))=-2/(nd)-
a3^2+a8^2+2a3a8=9(a3+a8)^2=9因为等差数列an的各项都是负数所以a3+a8=-3所以S10=(a1+a10)*10/2=5(a1+a10)=5(a3+a8)=5*(-3)=-
am=a1+(m-1)d=xan=a1+(n-1)d=y两式相减得(m-n)d=x-yd=(x-y)/(m-n)代入a1+(m-1)d=x得a1=(my-nx+x-y)/(m-n)所以ai=a1+(i
由“2的(n-1)次幂与an的乘积=a(n-1)”有a1=a2乘以2a2=a3乘以2的平方…………a(n-1)=an乘以2的(n-1)次幂所以:an=a(n-1)/2的(n-1)次幂=a(n-2)/[
Am=A1+(m-1)d=pAn=A1+(n-1)d=qA(m+n)=A1+(m+n-1)dAm+An=2A1+(m+n-2)d=A1+(m+n-1)d+(A1-d)=p+qA(m+n)=A1+(m+