If p, q, r are in A.P and x, y, z are in G.P., then prove that $x^{q - r} . y^{r - p} . z^{p - q} = 1$ .

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Solution

$p, q, r$ are in AP
Let the common difference be $d$
$⟹ d = q - p = r - q 2 d = r - p$
$x, y, z$ are in GP
Condition to be in GP
$y^{2} = x z y^{2 d} = x^{d} z^{d} \frac{y^{r - p}}{x^{- (r - q)} z^{- (q - p)}} = 1 x^{q - r} . y^{r - p} . z^{p - q} = 1$

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