What are the natural numbers x and y if C(x-1,y-1), C(x-1,y), C(x,y) are the terms of an A.P. and A(x,y), A(x,y+1), A(x+1,y+1) are the terms of a...

We'll apply the theorem of the arithmetical average of 3
consecutive terms of an arithmetical progression.

C(x-1,y)
= [C(x-1,y-1) + C(x,y)]/2

C(x-1,y) =
(x-1)!/y!(x-y-1)!

C(x-1,y-1) =
(x-1)!/(y-1)!(x-y)!

C(x,y) =
x!/y!(x-y)!

2(x-1)!/y!(x-y-1)! = (x-1)!/(y-1)!(x-y)! +
x!/y!(x-y)!

2(x-1)!/(y-1)!*y(x-y-1)! =
(x-1)!/(y-1)!*(x-y-1)!*(x-y) +
x!/(y-1)!*y*(x-y-1)!(x-y)

2(x-1)!*(x-y) = y*(x-1)! +
(x-1)!*x

We'll factorize by
(x-1)!:

2(x-1)!*(x-y) = (x-1)!(y +
x)

We'll simplify:

2x - 2y = y
+ x

x - 3y = 0

x =
3y(1)

We'll apply the theorem of the geometric average of 3
consecutive terms of an geometric progression.

A(x,y),
A(x,y+1), A(x+1,y+1)

[A(x,y+1)]^2 =
A(x,y)*A(x+1,y+1)

[x!/(y+1)!*(x-y-1)!]^2 = x!/y!(x-y)! *
(x+1)!/(y+1)!*(x-y)!

[x!/(y+1)!*(x-y-1)!]^2 =
(x!)^2*(x+1)/(y!)^2*(x-y-1)!^2*(x-y)^2

1/(y!)^2*(y+1)^2*(x-y-1)!^2
= (x+1)/(y!)^2*(x-y-1)!^2*(x-y)^2

1/(y+1) =
(x+1)/(x-y)^2

y = 1 => x =
3

The natural values of the numbers x and y
are: x = 3 and y = 1.