公比为3.a4=9求a1

lim(a1+a2+a3+.an)=a/(1-q),a2,a4,...是首项为aq,公比为q^2的等比数列,lim(a2+a4+.+a2n)=aq/(1-q^2),lim(a1+a2+a3+.an)/

-3再问：求过程。。。我也算得这个相当于上面a1/(1-(-1/3)^2)再除a1*(-1/3)/(1-(-1/3)^2)等于-3。。。可是选项中没这个。。。再答：就是把下面的每一项都提出来一个-1/

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

1.a4-a2=24a2+a3=6∴a3+a4=30a2+a3=a1q（1+q）=6a3+a4=a1q^2（1+q）=30两式相除得：q=5代入a1q(1+q)=6得a1=1/52.S3+S6=2S9

a1+a2+a3=a1(1+q+q^2)=1/2a4+a5+a6=a1q^3(1+q^2+q^2)=-41/q^3=(1/2)/(-4)q^3=-8q=-2

由a4a1＝q3＝648＝8可得q=2．

a1+a4=18a1+a1q^3=18a1(1+q^3)=18a1=18/(1+q^3)[q≠-1]a2+a3=12a1q+a1q^2=12a1(q+q^2)=12a1=12/(q+q^2)[q≠-1

∵an是等比数列∴a4=a1q³an=a1q^（n-1）8=q³q=2

a1(1+q+q^2)=1a1(q^3+q^4+q^5)=-2-->a1q^3(1+q+q^2)=-2两式相除,得：q^3=-2所以q=-2^(1/3)

(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)^(

a1+a2=a1+a1q=a1(1+q)=21a3+a4=a1q^2+a1q^3=a1q^2(1+q)=822式除1式得q^2=4q=±2分别代入1式得a1=2/3a1=-2(舍去q=-2)S8=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)

设数列an=a1*q^(n-1),所以a1+a4+a7+.+a97=a1+a1*q^3+a1*q^6+……a1q^96=a1(1+q^3+q^6+……+q^96)=5,a3+a6+a9+.+a99=a

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(

a4+a6=q^3*(a1+a3)5/4=q^3*10q^3=5/4÷10q^3=1/8q=1/2不懂可追问,有帮助请采纳,谢谢!

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