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
Given x, y, z are three terms in an arithmetic progression.
Considering x = a, y = a+d, z = a+2*d.
Using the given equation x*y*z = 5*(x+y+z)
a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)
=a*(a+d)*(a+2*d) = 5*(3*a+3*d) = 15*(a+d).
= a*(a+2*d) = 15.
Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d 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+2*d = 15.
x = 1, z = 15.
z-x = 14.