解: (1) An是等差数列,An=a1+(n-1)d Ab1,Ab2,Ab3...是等比, 因为b1=1,b2=2,b3=6,所以A1,A2,A6构成等比 A2^=A1*A6 即(A1+d)^2=A1*(A1+5d) 所以A1=d/3 q=A2/A1=(A1+d)/A1=4 (2)Abn=Ab1*q^(n-1)=A1*4^(n-1) (3)Abn=A1+(bn-1)d 所以A1+(bn-1)d=A1*4^(n-1) 又因为d=3A1,所以A1+(bn-1)*3A1=A1*4^(n-1) 所以bn=[2+4^(n-1)]/3