If a,b,c are in AP and x,y,z are in GP, prove that $x^{b - c} . y^{c - a} . z^{a - b} = 1$

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Solution

Given $a, b, c$ are in $A . P$
$⟹ 2 b = a + c$
And also $x, y, z$ are in $G . P$
$⟹ y = \sqrt{x z}$
$L H S = x^{b - c} . y^{c - a} . z^{a - b} = x^{b - c} . z^{a - b} . (\sqrt{x z})^{c - a} = x^{b - c} . z^{a - b} . x^{\frac{c - a}{2}} . z^{\frac{c - a}{2}} = x^{\frac{2 b - (a + c)}{2}} z^{\frac{2 b - (a + c)}{2}} = x^{0} . z^{0} = 1$

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If a b and c are in GP and $a^{1 / x} = b^{1 / y} = c^{1 / z}$ , prove that x, y and z are in AP.

x^{(b - c)} . y^{(c - a)} . z^{(a - b)}

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