Determine all pairs (a,b) of real numbers such that `10, a,b,ab` are in arithmetic progression .

Answer

To determine all pairs \((a, b)\) of real numbers such that \(10, a, b, ab\) are in arithmetic progression, we can follow these steps:

### Step 1: Understand the condition for arithmetic progression (AP)
For four numbers \(x_1, x_2, x_3, x_4\) to be in AP, the condition is:
\[ 2x_2 = x_1 + x_3 \quad \text{and} \quad 2x_3 = x_2 + x_4 \]
In our case, we have \(x_1 = 10\), \(x_2 = a\), \(x_3 = b\), and \(x_4 = ab\).

### Step 2: Set up the equations using the AP condition
From the first condition:
\[ 2a = 10 + b \quad \text{(1)} \]
From the second condition:
\[ 2b = a + ab \quad \text{(2)} \]

### Step 3: Solve equation (1) for \(b\)
Rearranging equation (1):
\[ b = 2a - 10 \quad \text{(3)} \]

### Step 4: Substitute \(b\) from equation (3) into equation (2)
Substituting \(b\) in equation (2):
\[ 2(2a - 10) = a + a(2a - 10) \]
Expanding both sides:
\[ 4a - 20 = a + 2a^2 - 10a \]
Rearranging gives:
\[ 2a^2 - 6a + 20 = 0 \]

### Step 5: Simplify the quadratic equation
Dividing the entire equation by 2:
\[ a^2 - 3a + 10 = 0 \]

### Step 6: Calculate the discriminant
The discriminant \(\Delta\) of the quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ \Delta = b^2 - 4ac \]
For our equation:
\[ \Delta = (-3)^2 - 4 \cdot 1 \cdot 10 = 9 - 40 = -31 \]
Since the discriminant is negative, there are no real solutions for \(a\).

### Conclusion
Thus, there are no pairs \((a, b)\) of real numbers such that \(10, a, b, ab\) are in arithmetic progression.

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