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Question
If a, b, c are in G.P. and a1/x = b1/y = c1/z, then xyz are in
Options
(a) AP
(b) GP
(c) HP
(d) none of these
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Solution
(a) AP
\[\text{ a, b and c are in G . P } . \]
\[ \therefore b^2 = ac\]
\[\text{ Taking log on both the sides }: \]
\[2\log b = \log a + \log c . . . . . . . . \left( i \right)\]
\[Now, a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z} \]
\[\text{ Taking \log on both the sides }: \]
\[\frac{\log a}{x} = \frac{\log b}{y} = \frac{\log c}{z} . . . . . . . . \left( ii \right)\]
\[\text{ Now, comparing } \left( i \right) \text{ and } \left( ii \right): \]
\[\frac{\log a}{x} = \frac{\log a + \log c}{2y} = \frac{\log c}{z}\]
\[ \Rightarrow \frac{\log a}{x} = \frac{\log a + \log c}{2y} \text{ and } \frac{\log a}{x} = \frac{\log c}{z}\]
\[ \Rightarrow \log a \left( 2y - x \right) = xlog c \text{ and } \frac{\log a}{\log c} = \frac{x}{z}\]
\[ \Rightarrow \frac{\log a}{\log c} = \frac{x}{\left( 2y - x \right)} \text{ and } \frac{\log a}{\log c} = \frac{x}{z}\]
\[ \Rightarrow \frac{x}{\left( 2y - x \right)} = \frac{x}{z}\]
\[ \Rightarrow 2y = x + z\]
\[\text{ Thus, x, y and z are in A . P } . \]
\[\]
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