If p, q, r are in AP and x, y, z are in GP, then $x^{q - r} . y^{r - p} . z^{p - q}$ is equal to

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Solution

The correct option is A 1
Let d be the common difference of AP and $R (\neq 0)$ , the common ratio of GP, then
q = p +d, r = p + 2d
and $y = x R, z = x R^{2}$
So, that q – r = –d, r –p = 2d, p – q = –d
$∴ x^{q - r} ∙ y^{r - p} ∙ z^{p - q} = x^{d} ∙ (x R)^{2 d} ∙ (x R^{2})^{- d}$
$= (x^{- d} ∙ x^{2 d} ∙ x^{- d}) (R^{2 d} ∙ R^{- 2 d})$
$= (x^{- d + 2 d - d}) ∙ (R^{2 d - 2 d})$
$= x^{0} ∙ R^{0} = 1 \times 1 = 1$

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