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, and z are in an arithmetic progression and that \(xyz = 5(x + y + z),\) we can start by simplifying the equation:

\(xyz = 5(3y)\) because x, y, and z are in an arithmetic progression.

This leads to xz = 15.

Now, we can explore the possible combinations for x and z, which are (3, 5) and (1, 15).

We also know that \(z - x > 4\) since \(y - x > 2.\)

Therefore, when we consider the combination (1, 15), we have:

\(z - x = 15 - 1 = 14.\)

Hence, \(z - x\) equals 14.

Answer 2

Given \(x, y, z\) are three terms in an arithmetic progression.
Let, \(x = a, y = a+d, z = a+2d\)
Using the given equation \(xyz = 5(x+y+z)\)
\(a(a+d)(a+2d) = 5(a+a+d+a+2d)\)
\(⇒a(a+d)(a+2d) = 5(3a+3d)\)
\(⇒a(a+d)(a+2d)= 15(a+d)\)
\(⇒ a(a+2d) = 15\)
Since all \(x, y, z\) are positive integers and \(y-x > 2\).
\(a, a+d, a+2d\) are integers.
The common difference is positive and greater than \(2\).
Among the different possibilities are :
\(a=1, a+2d = 5,\)
\(a, =3, a+2d = 5,\)
\(a = 5, a+2d = 3,\)
\(a=15, a+2d = 1\)
Hence the only possible case satisfying the condition is:
\(a = 1, a+2d = 15\)
\(x=1, z= 15\)
\(z-x =15-1= 14\)

So, the correct option is (C): \(14\)