我知道很简单,但我做不出来---///

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)=13/16a1*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