公比为3.a4=9求a1等比数列
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以下为解答.(2)由于bn=log2an(n∈N*),所以bn=n所以{bn}的前n项和:Sn=1+2+3+……+n=(n^2+n)/2
a3=a1+2d=a1+4a4=a1+3d=a1+6因为a1,a3,a4成等比数列,则a4/a3=a3/a1(a1+4)^2=a1(a1+6)解之,a1=-8则a2=a1+d=-8+2=-6
a4.a7=32,a5*a6=a4*a7=32a5+a6=12所以a5=4,a6=8或a5=8,a6=4从而1.q=a6/a5=2,a1=a5/q^4=4/16=1/42.q=a6/a5=1/2,a1
题目为:a1+a2+a3=48a2*a2=a1*a4an=a1+(n-1)d求解a1、d即可带入公式可把条件转化为3a1+3d=48(a1+d)(a1+d)=a1(a1+3d)解出a1=16d=0或:
a1*a2*a3=a1*(a1q)*(a1q^2)=27→a1^3*q^3=27→a1q=3=a2∵a2+a4=36∴a4=33a4/a2=q^2=11∴q=√11a1=√11/3好久不动数列了
由a4a1=q3=648=8可得q=2.
设等比数例{an}的公比为d,因为a1+a3=10所以a1+a1d^2=10,a1(1+d^2)=10,因为a4+a6=5/4,所以a1d^3+a1d^5=5/4,a1d^3(1+d^2)=5/4用a
An=A*q^(n-1)带进去算A(1+q^3)=18Aq(1+q)=12算出A=2q=2然后用等比数列和的算法算S8=510S8-30=答案!自己动笔算一算了
∵an是等比数列∴a4=a1q³an=a1q^(n-1)8=q³q=2
(a1+a3)q*q*q=a4+a6;即10*q*q*q=5/4;q=1/2;a1+a3=a1(1+q*q)=10;即a1(1+1/4)=10;a1=8;an=a1*q^(n-1)=8*(1/2)^(
易知(a2+d)^2=a2*(a2+4d)得:d=2a2所以(a1+a3+a5)/(a2+a4+a6)=(a2-d+a2+d+a2+3d)/(a2+a2+2d+a2+4d)=(a2+a2+a2+6a2
你的答案似乎不对,因为我做过这道题三遍.1.a1,a3,a5成等比则:a3^2=a1*a5又a1,a3,a5是等差数列{an}中的项则:a3=a1+2da5=a1+4d则有:(a1+2d)^2=a1(
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
a3*a4=a2*a5=1/2及a2+a5=9/4,得a2=2,a5=1/4,则1/4=a5=a2*q^3=2*q^3,得q=1/2,a1=4,则an=4*(1/2)^(n-1)=(1/2)(n-3)
a1+a4+a7+……+a28=a1(1+2^3+2^6+……+2^27)=a1[1-(2³)^10]/(1-2³)=100a3+a6+a9+……+a30.=a1(2^2+2^5+
因为a1*a2*a3=1/3^6,所以a2^3=1/3^6,所以a2=1/91/a2+1/a3+1/a4=(1+1/q+1/q^2)/a2=117,所以(1+1/q+1/q^2)=13解得q=1/3(
设公比为q则a3=a1q^2=7S3=a1+a2+a3=a1+a1q+a1q^2=a1(1+q+q^3)=7+a1(1+q)=21则a1=14/(1+q)则q=1或q=-1/2q=1,则a1=7q=-
a1+d=a2=b2=a1*d,a1*(d-1)=da1+3d=a4=b4=a1*d^3,a1*(d^3-1)=3d二式除一式得:d^2+d+1=3,d=1(舍)或d=-2,a1=2/3an=2/3-
因为{an}为等比数列所以an=a1*q^(n-1)a1*a5=a1*a1*q^4=16a1^2*q^4=16a1*q^2=±4所以a1=4/q^2①或a1=-4/q^2②a2+a4=a1*q+a1*
a4=a1*q^3a5=a2*q^3a6=a3*q^3∴(a4+a5+a6)/(a1+a2+a3)=q^3=56/7=8q=2公比为2