If $a, b, c$ are three consecutive terms of an A.P. and $x, y, z$ are three consecutive terms of a G.P. then prove that $x^{b - c} \times y^{c - a} \times z^{a - b} = 1$ .

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Solution

$a, b, c$ is three consecutive terms of an AP.
$∴$ Let $a, b, c$ be $a, a + d, a + 2 d$ respectively $\dots (1)$
$x, y, z$ is three consecutive terms of a GP.
$∴$ Assume $x, y, z$ as $x, x . r, x . r^{2}$ respectively ...(2)
To Prove: $x^{b - c}, y^{c - a}, z^{a - b} = 1$
Substituting (1) and (2) in L.H.S., we get
$\begin{matrix} L.H.S. = x^{a + d - a - 2 d} \times (x r)^{a + 2 d - a} \times {(x r^{2})}^{a - a - d} = (x)^{- d} \cdot (x)^{2 d} {(x r^{2})}^{- d} = \frac{1}{x^{4}} \times x^{2 d} \cdot r^{2 d} \times \frac{1}{x^{d} r^{2 d}} = 1 = R.H.S. \end{matrix}$
Hence proved.

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