If `a ,b ,c ,d` are in A.P. and `x , y , z` are in G.P., then show that `x^(b-c)doty^(c-a)dotz^(a-b)=1.`

Explanation

Let `D` is the common difference in the given A.P.

Then,

`b-a = D, c-b = D, d-c = D`

`:. b-c = -D, c-a = 2D,a-b = -D`

As, `x,y and z` are in G.P.

`:. y^2 = zx`

Now, `x^(b-c)*y^(c-a)*z^(a-b) = x^(-D)*y^(2D)*z^(-D)`

`= (1/(zx))^D*(y^2)^D`

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