(1)S1=a1=2a1-2 -> a1=2
S2=a1+a2=2a2-2 -> a2=4
S3=a1+a2+a3=2a3-2 -> a3=8
(2)Sn=2an-2 ①
S(n-1)=2a(n-1)-2 ②
①-②得an=2an-2a(n-1) -> an=2a(n-1) -> {an}为首项2,公比2的等比数列
所以 an=2*2^(n-1)=2^n
{bn}为等差数列,所以 bn=b1+(n-1)d=2+(n-1)d
b1,b3,b11成等比数列 -> b3^2=b1*b11 -> (2+2d)^2=2*(2+10d) -> d^2=3d -> d=0或d=3
又因为公差不为0,所以bn=2+3(n-1)=3n-1