Given `2B = A + C`
or `3B = pi ” or ” B = pi//3` (i)
Also a, b, c are in G.P. `rArr b^(2) = ac` (ii)
Now, `cos B = cos 60^(@) = (1)/(2) = (c^(2) a^(2) -b^(2))/(2ca)`
or `ca = c^(2) + a^(2) – b^(2)`
or `2b^(2) = c^(2) + a^(2)` [by using Eq. (ii)]
Hence, `a^(2), b^(2), c^(2)` are in A.P.

As angles `A,B and C` are in A.P.
`:. 2B = A+C`
Now, `A+B+C = 180 => B+2B+ 180`
`=>3B = 180 =>B= 60^@`
Now, `cos B = (a^2+c^2 – b^2)/(2ac)`
`=>cos 60^@ = (a^2+c^2 – b^2)/(2ac)`
`=>1/2 = (a^2+c^2 – b^2)/(2ac)`
`=>ac = a^2+c^2 – b^2->(1)`
Also, it is given that, `a,b and c` are in G.P.
`:. b^2 = ac`
Using, `b^2 = ac` in(1),
`=>b^2 = a^2+c^2 – b^2`
`=>2b^2 = a^2+c^2`
Hence, `a^2,b^2 and c^2` are in A.P.