Solution:

Since \(a, b, c\) are in Geometric Progression (G.P.), we can write: \[ b^2 = ac \] Now given: \[ a^{\tfrac{1}{x}} = b^{\tfrac{1}{y}} = c^{\tfrac{1}{z}} = k \quad (\text{say}) \] Taking logarithms: \[ \frac{1}{x}\ln a = \frac{1}{y}\ln b = \frac{1}{z}\ln c \] ⇒ \[ \ln a : \ln b : \ln c = \tfrac{1}{x} : \tfrac{1}{y} : \tfrac{1}{z} \] But since \(a, b, c\) are in G.P., \[ \ln a, \ln b, \ln c \] are in **A.P.** (because logarithm of G.P. terms forms an A.P.). Hence, the reciprocals \( \tfrac{1}{x}, \tfrac{1}{y}, \tfrac{1}{z} \) are also in A.P. Which implies: \[ x, y, z \ \text{are in Arithmetic Progression}. \]

✅ Correct Answer: (a) Arithmetic Progression