First slide

Geometric progression

Question

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

Moderate

A

1
B

2
C

-1
D

None of these

Solution

Let d be the common difference of A.P. and R (≠ 0), the common ratio of G.P., then
$q = p + d, r = p + 2 d$
and $y = xR, z = {xR}^{2}$
so that $q - r = - d, r - p = 2 d, p - q = - d$
$\begin{array}{l} ∴ x^{q - r} \cdot y^{r - p} \cdot z^{p - q} = x^{- d} \cdot (xR)^{2 d} \cdot {({xR}^{2})}^{- d} \\ = (x^{- d} \cdot x^{2 d} \cdot x^{- d}) (R^{2 d} \times R^{- 2 d}) \\ = (x^{- d + 2 d - d}) \cdot (R^{2 d - 2 d}) \\ = x^{0} \cdot R^{0} = 1 \times 1 = 1 \end{array}$

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