Question

Three positive integers x, y and z are in arithmetic progression. If \(y − x > 2 \) and \(xyz = 5(x + y + z),\) then \(z − x\) equals

• A. 8
• B. 10
• C. 14
• D. 12

Answer 1

The Correct Option is C

Given that \( x, y, z \) are in an arithmetic progression and: \[ xyz = 5(x + y + z) \] We are to find the value of \( z - x \), given that: \[ y - x > 2 \]

Step 1: Use Arithmetic Progression

Since \( x, y, z \) are in A.P., then: \[ y = \frac{x + z}{2} \quad \Rightarrow \quad x + y + z = 3y \] So the equation becomes: \[ xyz = 5(x + y + z) = 5 \cdot 3y = 15y \]

Rewrite the equation: \[ xyz = 15y \quad \Rightarrow \quad xz = 15 \tag{1} \] (Dividing both sides by \( y \), assuming \( y \ne 0 \))

Step 2: Find values of \( x \) and \( z \) such that \( xz = 15 \)

Possible positive integer factor pairs of 15:

• \( (1, 15) \)
• \( (3, 5) \)
• Note: \( x \) and \( z \) must allow an integer middle term \( y = \frac{x + z}{2} \)

Step 3: Check both cases for valid A.P. and condition \( y - x > 2 \)

Case 1: \( x = 1, z = 15 \)

\[ y = \frac{1 + 15}{2} = 8 \\ y - x = 8 - 1 = 7 > 2 \\ z - x = 15 - 1 = \boxed{14} \] ✅ Satisfies all conditions.

Case 2: \( x = 3, z = 5 \)

\[ y = \frac{3 + 5}{2} = 4 \\ y - x = 1 \quad \text{(Does not satisfy } y - x > 2 \text{)} \] ❌ Discard this case.

✅ Final Answer:

\[ z - x = \boxed{14} \]

Answer 2

Problem Statement: 

Given \( x, y, z \) are three terms in an arithmetic progression, and:

\[ xyz = 5(x + y + z) \]

Step 1: Represent terms in A.P.

Let:

\[ x = a, \quad y = a + d, \quad z = a + 2d \]

Step 2: Substitute into given equation

\[ a(a + d)(a + 2d) = 5[a + (a + d) + (a + 2d)] \Rightarrow a(a + d)(a + 2d) = 5(3a + 3d) \Rightarrow a(a + d)(a + 2d) = 15(a + d) \]

Step 3: Cancel common term \( (a + d) \)

Since \( a + d \neq 0 \), divide both sides:

\[ a(a + 2d) = 15 \]

Step 4: Use integer factor pairs of 15

We now look for integer values of \( a \) and \( a + 2d \) such that:

\[ a(a + 2d) = 15 \Rightarrow \text{Try factor pairs of 15: } (1, 15), (3, 5), (5, 3), (15, 1) \]

• \( a = 1, a + 2d = 15 \Rightarrow 2d = 14 \Rightarrow d = 7 \)
• \( a = 3, a + 2d = 5 \Rightarrow 2d = 2 \Rightarrow d = 1 \)
• \( a = 5, a + 2d = 3 \Rightarrow d = -1 \) (invalid, since \( d \) must be positive)
• \( a = 15, a + 2d = 1 \Rightarrow d = -7 \) (invalid)

Step 5: Check constraints

Given: \( y - x = d > 2 \), so only the case \( d = 7 \) satisfies this.

Hence, the only valid case is:

\[ a = 1, \quad d = 7 \Rightarrow x = 1, y = 8, z = 15 \]

\[ z - x = 15 - 1 = \boxed{14} \]

✅ Final Answer:

Option (C): \( \boxed{14} \)