f(x)=ax+b,不达原点,则有b0
f(2)=2a+b=t,b=t-2a
f(f(2))=at+b
2,f(2),f(f(2))成等比数列,则有：t^2=2*(at+b)
t^2-2at-2b=0
t^2-2at-2(t-2a)=0
t^2-2t(a+1)+4a=0
(t-2a)(t-2)=0
得：t=2a,or 2
t=2a时,b=0,不符.
因此只能为t=2.即f(2)=2