Since $10, a, b, ab$ is an arithmetic progression, we have $a-10 = b-a = ab-b$.
We can rewrite these equations as $a-10 = b-a$ and $b-a = ab-b$.
From the first equation, we have $2a = 10 + b$, so $b = 2a - 10$.
Substituting this into the second equation, we get $2a - 10 - a = a(2a - 10) - (2a - 10)$,
which simplifies to $a - 10 = a(2a - 10) - 2a + 10$.
Expanding and rearranging, we have $a^2 - 12a + 20 = 0$.
Factoring, we get $(a-2)(a-10) = 0$, so $a = 2$ or $a = 10$.
If $a = 2$, then $b = 2a - 10 = 2(2) - 10 = -6$.
If $a = 10$, then $b = 2a - 10 = 2(10) - 10 = 10$.
Therefore, there are $\boxed{2}$ pairs $(a,b)$ that satisfy the condition.The answer is: 2
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Determine the number of pairs $(a,b)$ of real numbers such that $10, a, b, ab$ is an arithmetic progression.

### Answer:
