2(a(n+1)+d)=an+d
2a(n+1)=an-d
故d=-1
2（a(n+1)-1)=an-1
a(n+1)-1=1/2(an-1)
于是數(shù)列{an-1}是首項(xiàng)a1-1=3-1=2,公比q=1/2的等比數(shù)列
故an-1=2*(1/2)^(n-1)=(1/2)^(n-2)
an=(1/2)^(n-2)+1