Find all triples of three-digit positive integers $x < y < z$ with $x,y,z$ in arithmetic progression and $x, y, z + 1000$ in geometric progression.

[i]For this problem, you may use calculators or computers to gain an intuition about how to solve the problem. However, your final submission should include mathematical derivations or proofs and should not be a solution by exhaustive search.[/i]
1. **Arithmetic Progression Condition**: Given \( x, y, z \) are in arithmetic progression, we have:
   \[
   y - x = z - y \implies 2y = x + z \implies z = 2y - x \quad \text{(1)}
   \]

2. **Geometric Progression Condition**: Given \( x, y, z+1000 \) are in geometric progression, we have:
   \[
   \frac{y}{x} = \frac{z+1000}{y} \implies y^2 = x(z + 1000) \quad \text{(2)}
   \]

3. **Substitute \( z \) from (1) into (2)**:
   \[
   y^2 = x(2y - x + 1000)
   \]
   Simplify the equation:
   \[
   y^2 = 2xy - x^2 + 1000x
   \]
   Rearrange to form a quadratic equation:
   \[
   y^2 - 2xy + x^2 = 1000x
   \]
   Recognize the left-hand side as a perfect square:
   \[
   (y - x)^2 = 1000x \quad \text{(3)}
   \]

4. **Solve for \( y \)**: Since \( y > x \), \( y - x \) is positive, and we can take the square root of both sides:
   \[
   y - x = \sqrt{1000x} \implies y = x + \sqrt{1000x} \quad \text{(4)}
   \]

5. **Ensure \( y \) is an integer**: For \( y \) to be an integer, \( \sqrt{1000x} \) must be an integer. Let \( \sqrt{1000x} = 10a \) where \( a \) is an integer:
   \[
   1000x = (10a)^2 \implies x = 10a^2
   \]

6. **Determine \( y \)**: Substitute \( x = 10a^2 \) into (4):
   \[
   y = 10a^2 + 10a(a) = 10a(a + 10)
   \]

7. **Check the range for \( x \) and \( y \)**: Since \( x \) and \( y \) are three-digit numbers:
   \[
   100 \leq 10a^2 < 1000 \implies 10 \leq a^2 < 100 \implies 4 \leq a \leq 9
   \]
   Calculate \( x \) and \( y \) for \( a = 4, 5, 6, 7, 8, 9 \):
   - \( a = 4 \): \( x = 160 \), \( y = 560 \)
   - \( a = 5 \): \( x = 250 \), \( y = 750 \)
   - \( a = 6 \): \( x = 360 \), \( y = 960 \)
   - \( a = 7 \): \( x = 490 \), \( y = 1190 \) (not a three-digit number)
   - \( a = 8 \): \( x = 640 \), \( y = 1440 \) (not a three-digit number)
   - \( a = 9 \): \( x = 810 \), \( y = 1710 \) (not a three-digit number)

8. **Determine \( z \)**: Using \( z = 2y - x \):
   - For \( (160, 560) \): \( z = 2(560) - 160 = 960 \)
   - For \( (250, 750) \): \( z = 2(750) - 250 = 1250 \) (not a three-digit number)
   - For \( (360, 960) \): \( z = 2(960) - 360 = 1560 \) (not a three-digit number)

9. **Verify the solution**: Check if \( (160, 560, 960) \) satisfies the geometric progression condition:
   \[
   \frac{560}{160} = \frac{960 + 1000}{560} \implies \frac{7}{2} = \frac{1960}{560} \implies \frac{7}{2} = \frac{7}{2}
   \]
   The condition is satisfied.

The final answer is \(\boxed{(160, 560, 960)}\)