Solution: Let the first term of series an be a 1 and the tolerance be d, and the general term formula is as follows: A5 = A 1+4d, and A13 = a1+12d.

So a5+a 13=34 is 2a 1+ 16d=34, which means a 1+8d= 17.

S3=9 means 3a 1+3d=9, that is, a 1+d=3, and the simultaneous solution is: a 1= 1 d=2, so the general formula of series an is an=2n- 1.

So the sequence b1= a1/a1+t =1/(2t+1).

b2=a2/a2+t=3/(2t+3)

bm=(2m- 1)/(2t+2m- 1)

Suppose B 1, B2 and BM are arithmetic progression, and the tolerance d = B2-b1= 3/(2t+3)-1(2t-1) = 4t/(2t+3) (2t+/kloc-.

Then BM = B2+D = 3/(2t+3)+4t/(2t+3) (2t+1) = (10t+1)/(2t+3) (2t+1).

Since bm=bm is an identity, there is 2m-1=10t+1; (2t+2m-1) = (2t+3) (2t+1) and the quadratic equation about t is obtained. The discriminant of the equation is less than zero, so it is meaningless.

There is no t that makes b 1, b2, bm form arithmetic progression.