if x,y,x are in GP and a^x,b^y,c^z are equal then a,c,b are in i)AP ii)GP iii)HP

Take x.,y,z as any gp, say 3, 9,27 respectively. Now to a^x=b^y=c^z take a,b, cas any descending  gp say 27, 9,3.  Hwnce the correct option is GP. 

Take x = 2,y = 6,z = 18 . Now take a = 2^9,b = 2^3,c = 2.It satisfies a^x=b^y=c^z .But it's not GP

Let us take x, y, z to be k/r, k, kr respectively.
a^x = b^y = c^y implies a^(k/r) = b^k = c^(kr)
If we take logarithm for all we get,
(k/r) loga = k logb = kr logc
(1/r) loga = logb = r logc
a^(1/r) = b = c^r
Clearly we can observe that a, c, b follow No Progression among AP, GP and HP.

GP

Let us take x, y, z to be k/r, k, kr respectively.
a^x = b^y = c^y implies a^(k/r) = b^k = c^(kr)
If we take logarithm for all we get,
(k/r) loga = k logb = kr logc
(1/r) loga = logb = r logc
a^(1/r) = b = c^r

Clearly we can observe that a, c, b follow No Progression among AP, GP and HP.

Hope this helps!