a(n)=(1/2)q^(n-1),
由于x^2-mx+2=0的兩個根的乘積為2, x^2-nx+2=0的兩個根的乘積也為2.
這樣,只有,
2=a(1)a(4)=a(2)a(3)=(1/2)^2q^3, q=2.

a(1)+a(4)=m=(1/2)+(1/2)q^3=(1/2)[1+q^3],

a(2)+a(3)=n=(1/2)q+(1/2)q^2=(1/2)q[1+q].
m/n = (1+q^3)/[q(1+q)]=9/[2*3]=3/2
或
a(1)+a(4)=n=(1/2)+(1/2)q^3=(1/2)[1+q^3],

a(2)+a(3)=m=(1/2)q+(1/2)q^2=(1/2)q[1+q].
m/n = q(1+q)/(1+q^3)=(2*3)/9=2/3
綜合,有
m/n = 3/2或2/3