∵am-1am+1-2am=0,由等比数列的性质可得,am2?2am=0∵am≠0∴am=2∵T2m-1=a1a2…a2m-1=(a1a2m-1)?(a2a2m-2)…am=am2m?2am=am2m?1=22m-1=128∴2m-1=7∴m=4故答案为4