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Let's analyze the problem step by step:

### Given Information:
- \( OAN \), \( OMB \), and \( APB \) are straight lines.
- \( AN = 2OA \), which means the vector \( \overrightarrow{ON} = 2\overrightarrow{OA} \).
- \( M \) is the midpoint of \( OB \), so \( \overrightarrow{OM} = \frac{1}{2} \overrightarrow{OB} \).
- \( \overrightarrow{OA} = \mathbf{a} \) and \( \overrightarrow{OB} = \mathbf{b} \), where \( \mathbf{a} \) and \( \mathbf{b} \) are vectors.
- \( \overrightarrow{AP} = k \overrightarrow{AB} \), where \( k \) is a scalar.
- \( MPN \) is a straight line, which implies the points \( M \), \( P \), and \( N \) are collinear.

### Objective:
We need to find the scalar \( k \), using the fact that \( MPN \) is a straight line.

### Steps:
1. **Express vectors in terms of \( \mathbf{a} \) and \( \mathbf{b} \):**
   - \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \).
   - \( \overrightarrow{AP} = k(\mathbf{b} - \mathbf{a}) \), because \( \overrightarrow{AP} = k \overrightarrow{AB} \).

2. **Express vector \( \overrightarrow{PN} \):**
   - \( \overrightarrow{PN} = \overrightarrow{ON} - \overrightarrow{OP} = 2\mathbf{a} - (\mathbf{a} + k(\mathbf{b} - \mathbf{a})) \).
   - Simplifying \( \overrightarrow{PN} \):  
     \[
     \overrightarrow{PN} = 2\mathbf{a} - \mathbf{a} - k(\mathbf{b} - \mathbf{a}) = (1 + k)\mathbf{a} - k\mathbf{b}.
     \]

3. **Vector \( \overrightarrow{PM} \):**
   - \( \overrightarrow{PM} = \overrightarrow{OM} - \overrightarrow{OP} = \frac{1}{2} \mathbf{b} - (\mathbf{a} + k(\mathbf{b} - \mathbf{a})) \).
   - Simplifying \( \overrightarrow{PM} \):  
     \[
     \overrightarrow{PM} = \frac{1}{2} \mathbf{b} - \mathbf{a} - k(\mathbf{b} - \mathbf{a}) = (-1 - k) \mathbf{a} + \left( \frac{1}{2} - k \right) \mathbf{b}.
     \]

4. **Collinearity condition:**
   - Since \( M \), \( P \), and \( N \) are collinear, \( \overrightarrow{PM} \) and \( \overrightarrow{PN} \) must be parallel vectors.
   - This means there must exist a scalar \( \lambda \) such that \( \overrightarrow{PN} = \lambda \overrightarrow{PM} \).

5. **Setting up the system:**
   \[
   (1 + k) \mathbf{a} - k \mathbf{b} = \lambda [(-1 - k) \mathbf{a} + (\frac{1}{2} - k) \mathbf{b}]
   \]
   By equating coefficients of \( \mathbf{a} \) and \( \mathbf{b} \), we can solve for \( k \).

I'll now proceed to solve this system for \( k \).The solutions for \( k \) are:

- \( k = -1 \)
- \( k = 0.25 \)

Given that the problem typically looks for a positive scalar \( k \), we select \( k = 0.25 \).

Thus, the value of \( k \) is \( \boxed{0.25} \).

Let me know if you'd like more details, and here are 5 related questions to deepen your understanding:

1. How would the value of \( k \) change if \( M \) were not the midpoint of \( OB \)?
2. What are the implications if the condition \( MPN \) being a straight line is removed?
3. Can you derive the equation for \( k \) if the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) were given numerically?
4. How would the solution change if the ratio \( AN : OA \) was different?
5. What happens to the geometry of the system if \( k \) takes a negative value?

**Tip:** In vector problems involving collinearity, always check for proportionality conditions between vectors for key insights into the system's behavior.

Given that OAN, OMB, and APB are straight lines, AN = 2OA, M is the midpoint of OB, OA = a, OB = b, AP = kAB where k is a scalar quantity. Given that MPN is a straight line, find the value of k.

This problem explores the use of vectors to find a scalar k when given conditions about collinearity and midpoints. Learn how to apply vector addition, scalar multiplication, and the collinearity condition to solve for k in this advanced geometry problem, suitable for students in Grades 11-12.

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