a3=a1+2d
q9=a1+8d
a1,a3,a9成等比数列
所以(a1+2d)^2=a1*(a1+8d)
a1^2+4a1d+4d^2=a1^2+8a1d
d^2=a1d
d≠0
d=a1
所以a1+a3+a9=a1+(a1+2d)+(a1+8d)=a1+3a1+9a1=13a1
a2+a4+a10=(a1+d)+(a1+3d)+(a1+9d)=2a1+4a1+10a1=16a1
所以(a1+a3+a9)/(a2+a4+a10)=13/16
a1*a3*a5*...*a(2n-1)*a(2n+1)=100 ...(1)
a2*a4*a6*...*a(2n)
=a1*q*a3*q*...*a(2n-1)*q
=a1*a3*...*a(2n-1)*q^n=120 .(2)
(2)式除以(1)式得
:q^n/a(2n+1) = 120/100 = 6/5
所以q^n/(a(n+1)*q^n)=1/a(n+1)=6/5
所以a(n+1)=5/6