∵a_m-1a_m+1-2a_m=0，
由等比数列的性质可得，am2−2am＝0
∵a_m≠0
∴a_m=2
∵T_2m-1=a₁a₂…a_2m-1=（a₁a_2m-1）•（a₂a_2m-2）…a_m
=am2m−2am=am2m−1=2^2m-1=128
∴2m-1=7
∴m=4
故答案为4