# 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 G.P. ?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.