Given that \( x, y, z \) are in an arithmetic progression and \( xyz = 5(x + y + z) \). We need to find \( z - x \) given that \( y - x > 2 \).
For an arithmetic progression \( x, y, z \), we have \( y = \frac{x + z}{2} \), which implies \( x + y + z = 3y \). Substituting this into the given equation:
\[ xyz = 5(x + y + z) = 5(3y) = 15y \] To simplify further, we rewrite the equation as:
\[ xyz = 15y \quad \Rightarrow \quad xz = 15 \quad (1) \] This is derived by dividing both sides by \( y \), assuming \( y e 0 \).
We need to find integer pairs \( (x, z) \) such that \( xz = 15 \) and \( y = \frac{x + z}{2} \) is also an integer. The positive integer factor pairs of 15 are:
We will now check each pair against the condition \( y - x > 2 \).
Case 1: \( x = 1, z = 15 \)
Calculate \( y \): \( y = \frac{1 + 15}{2} = 8 \). Check the condition \( y - x > 2 \): \( 8 - 1 = 7 \). Since \( 7>2 \), this case is valid. Calculate \( z - x \): \( z - x = 15 - 1 = \boxed{14} \). This case satisfies all conditions.
Case 2: \( x = 3, z = 5 \)
Calculate \( y \): \( y = \frac{3 + 5}{2} = 4 \). Check the condition \( y - x > 2 \): \( 4 - 3 = 1 \). Since \( 1 \) is not greater than \( 2 \), this case is invalid and must be discarded.
The only valid solution yields \( z - x = \boxed{14} \).