If a, b, c are in G. P. and $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$, then x, y, z:

Question

Accepted Answer

To solve the problem, we need to analyze the given conditions and determine why Option 1 is the correct answer.

### Explanation:

1. **Given Conditions:**
   - $ a, b, c $ are in Geometric Progression (G.P.).
   - $ a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} $.

2. **Understanding the Conditions:**
   - Since $ a, b, c $ are in G.P., we have the relation $ b^2 = ac $.
   - The condition $ a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k $ (say) implies:
     - $ a = k^x $
     - $ b = k^y $
     - $ c = k^z $

3. **Substituting in the G.P. Condition:**
   - From $ b^2 = ac $, substituting the expressions for $ a, b, $ and $ c $, we get:
     $$
     (k^y)^2 = k^x \cdot k^z
     $$
     $$
     k^{2y} = k^{x+z}
     $$
   - Since the bases are the same, we equate the exponents:
     $$
     2y = x + z
     $$

4. **Conclusion:**
   - The equation $ 2y = x + z $ is the condition for $ x, y, z $ to be in Arithmetic Progression (A.P.), because in an A.P., the middle term is the average of the other two terms, i.e., $ 2y = x + z $.

### Why Other Options are Incorrect:

- **Option 2 (G.P.):** For $ x, y, z $ to be in G.P., the condition would be $ y^2 = xz $. However, we derived $ 2y = x + z $, which does not satisfy the condition for a geometric progression.

- **Option 3 (H.P.):** For $ x, y, z $ to be in Harmonic Progression (H.P.), the reciprocals should be in A.P., i.e., $ \frac{2}{y} = \frac{1}{x} + \frac{1}{z} $. This is not equivalent to the derived condition $ 2y = x + z $.

- **Option 4 (Relation $2x^2 + y^2 - z = 0$):** This is a specific algebraic relation that does not match the derived condition $ 2y = x + z $. There is no basis from the given conditions to derive such a relation.

Thus, the correct answer is Option 1, as $ x, y, z $ are in Arithmetic Progression.