Question

If a1/x=b1/y=c1/z{{a}^{1/x}}={{b}^{1/y}}={{c}^{1/z}}and a, b, ca,\ b,\ c are in g.P., then x, y, zx,\ y,\ z will be in

Answer 1

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Let a^1/x=b^1/y=c^1/z=m .soa=m^x,b=m^y, c=m^z. Now a b c are in gp so b^2=ac. This imply m^2y=m^z.m^x so comparing we get 2y=x+z. Hence a b c are in ap thank☺☺☺☺

Answer 2

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if a, b, c are in GP and a^1/x=b^1/y= c^1/z are in Ap