∫a 1 = 3，b 1= 1，a2=b2，3a5=b3，
∴a2=3+d=q=b2，
3a5=3(3+4d)=q2=b3，
Solve the equation to get q=3 or q=9.
When q=3, d=0, which does not meet the meaning of the question, so it is discarded;
When q=9, d = 6.
an=3+(n- 1)×6=6n-3，bn=qn- 1=9n- 1。
∫an = 3 logubn+v = logu(93n？ 3)+v，
∴6n-3-v=logu(93n？ 3),
When n= 1 and 3-v=logu 1=0,
∴v=3.
When n=2, 12-3-3=logu93,
u6=93，u=3，
∴u+v=6.
So the answer is: 6.