if a,b,c are in a.p and x,y,z are in g.p. then prove that x^b-c x y ^c-a x z^a-b = 1 descriptive form???????

x(b-c) .y (c-a).z(a-b) = 1

As a , b ,c are in AP , so 2b = a +c  (1)
x , y , z are in GP , so y2 = xz  (2)

Hence x(b-c) .y (c-a).z(a-b)  = x(b-c) .(xz) (c-a)/2.z(a-b)   ( using (2) )
= x (b-c) .x(c-a)/2 .z (c-a)/2 .z (a-b)
= xb-c +c-a2.za-b +c-a2
= x (2b -c -a)/2 .z (a + c -2b)/2
But 2b -a -c = a+c -2b = 0  from (1)
So x0 .z0 = 1.1 = 1 ( hence proved )

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