解:∵1,a1,a2,a3,…,an,2成等比数列,
∴a1an=a2an-1=a3an-2=…=akan-k+1=…=1×2=2.
∴An2=(a1an)(a2an-1)(a3an-2)…(an-1a2)(ana1)=(1×2)n=2n.
∴An=.
又∵1,b1,b2,b3,…,bn,2成等差数列,
∴b1+bn=1+2=3. ∴Bn==n.