2倍的根号5
20
解: 设a(n)=q*a(n-1)=q^(n-1)*a(1)
a1*a2*a3=q^3*a(1)=5
a7*a8*a9=q^24*a(1)=10
两式相除得到 q^21=2
又a4*a5*a6=q^15*a(1)=....
a1*a2*a3=5a7*a8*a9=5a1*a2*a3*(q^18)=10
q^18=1/5
a4*a5*a6=a1*a2*a3*q^9=10*(1/5)^(-1/2)
2500
a1*a3=a2^2 a1*a2*a3=a2^3
a7*a9=a8^2 a7*a8*a9=a8^3
a2*a8=a5^2
a4*a6=a5^2
a4*a5*a6=a5^3=50^2=2500