Determine all pairs $ (a,b)$ of real numbers such that $ 10, a, b, ab$ is an arithmetic progression.
To determine all pairs \((a, b)\) of real numbers such that \(10, a, b, ab\) form an arithmetic progression, we need to use the properties of arithmetic progressions. 

1. **Identify the common difference:**
   Let the common difference of the arithmetic progression be \(d\). Then we have:
   \[
   a = 10 + d
   \]
   \[
   b = 10 + 2d
   \]
   \[
   ab = 10 + 3d
   \]

2. **Express \(ab\) in terms of \(a\) and \(b\):**
   From the above, we know:
   \[
   ab = (10 + d)(10 + 2d)
   \]

3. **Set up the equation:**
   Since \(ab\) is also given by \(10 + 3d\), we equate the two expressions:
   \[
   (10 + d)(10 + 2d) = 10 + 3d
   \]

4. **Expand and simplify:**
   \[
   100 + 20d + 10d + 2d^2 = 10 + 3d
   \]
   \[
   2d^2 + 30d + 100 = 10 + 3d
   \]
   \[
   2d^2 + 27d + 90 = 0
   \]

5. **Solve the quadratic equation:**
   Using the quadratic formula \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 27\), and \(c = 90\):
   \[
   d = \frac{-27 \pm \sqrt{27^2 - 4 \cdot 2 \cdot 90}}{2 \cdot 2}
   \]
   \[
   d = \frac{-27 \pm \sqrt{729 - 720}}{4}
   \]
   \[
   d = \frac{-27 \pm \sqrt{9}}{4}
   \]
   \[
   d = \frac{-27 \pm 3}{4}
   \]
   \[
   d = \frac{-24}{4} \quad \text{or} \quad d = \frac{-30}{4}
   \]
   \[
   d = -6 \quad \text{or} \quad d = -7.5
   \]

6. **Find corresponding \(a\) and \(b\):**
   - For \(d = -6\):
     \[
     a = 10 + d = 10 - 6 = 4
     \]
     \[
     b = 10 + 2d = 10 + 2(-6) = 10 - 12 = -2
     \]
   - For \(d = -7.5\):
     \[
     a = 10 + d = 10 - 7.5 = 2.5
     \]
     \[
     b = 10 + 2d = 10 + 2(-7.5) = 10 - 15 = -5
     \]

Thus, the pairs \((a, b)\) are \((4, -2)\) and \((2.5, -5)\).

The final answer is \(\boxed{(4, -2) \text{ or } (2.5, -5)}\).