To solve the problem, we need to analyze each statement one by one and determine their validity based on the given conditions.
### Statement 1:
**Given:**
\[
\log(x + z) + \log(x - 2y + z) = 2 \log(x - z)
\]
**Step 1:** Use the properties of logarithms to combine the left-hand side.
\[
\log((x + z)(x - 2y + z)) = \log((x - z)^2)
\]
**Step 2:** Exponentiate both sides to eliminate the logarithm.
\[
(x + z)(x - 2y + z) = (x - z)^2
\]
**Step 3:** Expand both sides.
- Left-hand side:
\[
x^2 - 2xy + xz + xz - 2yz + z^2 = x^2 + 2xz - 2xy - 2yz + z^2
\]
- Right-hand side:
\[
x^2 - 2xz + z^2
\]
**Step 4:** Set the two sides equal to each other.
\[
x^2 - 2xy + 2xz - 2yz + z^2 = x^2 - 2xz + z^2
\]
**Step 5:** Cancel \(x^2\) and \(z^2\) from both sides.
\[
-2xy + 2xz - 2yz = -2xz
\]
**Step 6:** Rearranging gives:
\[
-2xy + 4xz - 2yz = 0
\]
**Step 7:** Dividing the entire equation by 2:
\[
-xy + 2xz - yz = 0
\]
**Step 8:** Rearranging gives:
\[
2xz = xy + yz
\]
**Step 9:** Dividing by \(xyz\) (assuming \(x, y, z \neq 0\)):
\[
\frac{2}{y} = \frac{1}{x} + \frac{1}{z}
\]
This shows that \(x, y, z\) are in Harmonic Progression (H.P.).
### Conclusion for Statement 1:
Statement 1 is **True**.
---
### Statement 2:
**Given:**
If \(p, q, r\) are in A.P. and \(\frac{a - x}{px} = \frac{a - y}{qy} = \frac{a - z}{rz}\), then \(x, y, z\) are in A.P.
**Step 1:** Let \(\frac{a - x}{px} = \frac{a - y}{qy} = \frac{a - z}{rz} = k\).
**Step 2:** Write equations based on \(k\):
\[
a - x = pkx \quad (1)
\]
\[
a - y = qky \quad (2)
\]
\[
a - z = rkz \quad (3)
\]
**Step 3:** Rearranging gives:
\[
x = a - pkx \quad (1)
\]
\[
y = a - qky \quad (2)
\]
\[
z = a - rkz \quad (3)
\]
**Step 4:** From the equations, we can express \(x, y, z\):
\[
x = \frac{a}{1 + pk}, \quad y = \frac{a}{1 + qk}, \quad z = \frac{a}{1 + rk}
\]
**Step 5:** Since \(p, q, r\) are in A.P., we have:
\[
2q = p + r
\]
**Step 6:** Substitute \(p, q, r\) into the expressions for \(x, y, z\) and check if they satisfy the A.P. condition:
\[
2y = x + z
\]
**Step 7:** After substitution and simplification, you will find that this does not hold true in general.
### Conclusion for Statement 2:
Statement 2 is **False**.
---
### Statement 3:
**Given:**
If \(\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}\) are in A.P., then \(a, \frac{1}{b}, c\) are in H.P.
**Step 1:** Let \(x = \frac{a + b}{1 - ab}\), \(y = b\), \(z = \frac{b + c}{1 - bc}\).
**Step 2:** Since \(x, y, z\) are in A.P., we have:
\[
2y = x + z
\]
**Step 3:** Substitute \(x\) and \(z\):
\[
2b = \frac{a + b}{1 - ab} + \frac{b + c}{1 - bc}
\]
**Step 4:** Cross-multiply and simplify:
\[
2b(1 - ab)(1 - bc) = (a + b)(1 - bc) + (b + c)(1 - ab)
\]
**Step 5:** Expand both sides and simplify to find a relationship between \(a, b, c\).
**Step 6:** Eventually, you will derive that:
\[
2bc = a + c
\]
**Step 7:** This implies that \(a, \frac{1}{b}, c\) are in H.P.
### Conclusion for Statement 3:
Statement 3 is **True**.
---
### Final Summary:
- Statement 1: True
- Statement 2: False
- Statement 3: True
