证明：设{a_n}中首项为a1,公差为d．
∵lga1,lga2,lga4成等差数列∴2lga2=lga1+lga4∴a22=a1?a4．
即（a1+d）2=a1（a1+3d）∴d=0或d=a1．
当d=0时,an=a1,bn=1a2n=1a1,∴bn+1bn=1,∴{b_n}为等比数列；
当d=a1时,an=na1,bn=1a2n=12na1,∴bn+1bn=12,∴{b_n}为等比数列．
综上可知{b_n}为等比数列．
打字不易,