Solution:
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
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