(Ⅰ)设{an}中首项为a1,公差为d.lga1,lga2,lga4成等差数列,把11和d代入求得d,进而分别当d=0,整理可得 bn+1•bn=1,进而判断出{bn}为等比数列;进而讨论d=a1时,整理即可判断出{bn}为等比数列.
(Ⅱ)把第一问所求结论分别代入即可求出数列{an}的首项a1和公差d.
【解析】
(Ⅰ)证明:设{an}中首项为a1,公差为d.
∵lga1,lga2,lga4成等差数列∴2lga2=lga1•lg44
∴a22=a1•a4.
即(a1+d)2=a1(a1+3d)∴d=0或d=a1.
当d=0时,an=a1,bn=,∴=1,∴{bn}为等比数列;
当d=a1时,an=na1,bn=,∴,∴{bn}为等比数列.
综上可知{bn}为等比数列.
(Ⅱ)当d=0时,=,所以a1=;
当d=a1时,=,
所以,故a1=3=d.
综上可知 或 .