if a b c are in ap and in gp find values of a^a-b+b^b-c+c^c-a

since a,b,c are in ap
so b= (a+c)/2......... (1)
and a,b,c are also in GP
so b^2= ac............ (2)
by putting (1) in (2), we get
(a+c)^2= 4ac
(a-c)^2=0 => a=c
and b= (a+a)/2= a
so a=b=c
now a^(a-b) +b^(b-c) + c^(c-a)
=> a^0+a^0+a^0= 1+1+1=3
so the value is 3 …....... [Ans]