在等比数列An中,(1)若a1=256,a9=1,求q和a12 (2)若a3*a5=18,a4*a8=72,求q

a1>a9
0<q<1

a9=a1q^8
1=256*q^8
q^8=1/256
q^4=1/16
q^2=1/4
q^2=±1/2
q=1/2

a12=a1q^11
=256*(1/2)^11
=2^8*2^-11
=2^-3
=1/8


a3*a5=18,
(a4)^2=18

a4*a8=72
(a6)^2=72

q^2=a6/a4
q^4=(a6)^2/(a4)^2
q^4=72/18
q^4=4
q^2=2
q=±√2

解由q^5=a9/a4=128/4=32=2^5
故q=2
则an=a4(q)^(n-4)=4×2^(n-4)=2^(n-2)
故a12=2^10=1024

a9/a6=q^3=8,所以q=2, a12=a9*q^3=-128,
a5=-1,,a4=-1/2,,a3=-1/4,,a2=-1/8,,a1=-1/16.
s5=-1-1/2-1/4-1/8-1/16=-31/16

a6=a1*q^5=-2
a9=a1*q^8=-16
q=2，a1=-1/16
a12=a1*q^11=-1/16*2^11=-2^7=-128
S5=a1+a2+a3+a4+a5=-1/16*(1+2+4+8+16)=-31/16