The $p^{t h}, q^{t h}$ and $r^{t h}$ term of an $A . P$ as well as those of $G . P$ are $a, b, c$ respectively then prove that $(a^{b - c}) (b^{c - a}) (c^{a - b}) = 1$

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Solution

Let n be the common difference of A.P.
$n (q - p) = b - a n (r - q) = c - b$
Let m be the common ratio of G.P.
$m^{q - p} = \frac{b}{a} m^{r - q} = \frac{c}{b} (\frac{b}{a})^{r - q} = (\frac{c}{b})^{2 - p} (\frac{b}{a})^{\frac{c - b}{n}} = (\frac{c}{b})^{\frac{b - a}{n}} \frac{b^{c - b}}{a^{c - b}} = \frac{c^{b - a}}{b^{b - a}} b^{c - a} = (a^{c - b}) (c^{b - a}) b^{c - a} = (a^{b - c}) (c^{a - b})^{- 1} (a^{b - c}) (b^{c - a}) (c^{a - b}) = 1$

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