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Question
Let x be the A.M. and y, z be two G.M.s between two positive numbers. Then, \[\frac{y^3 + z^3}{xyz}\] is equal to
Options
(a) 1
(b) 2
(c) \[\frac{1}{2}\]
(d) none of these
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Solution
(b) 2
\[\text{ Let the two numbers be a and b } . \]
\[\text{ a, x and b are in A . P }. \]
\[ \therefore 2x = a + b (i)\]
\[\text{ Also, a, y, z and b are in G . P } . \]
\[ \therefore \frac{y}{a} = \frac{z}{y} = \frac{b}{z}\]
\[ \Rightarrow y^2 = az , yz = ab, z^2 = by (ii)\]
\[\text{ Now }, \frac{y^3 + z^3}{xyz}\]
\[ = \frac{y^2}{xz} + \frac{z^2}{xy} \]
\[ = \frac{1}{x}\left( \frac{y^2}{z} + \frac{z^2}{y} \right)\]
\[ = \frac{1}{x}\left( \frac{az}{z} + \frac{by}{y} \right) \left[ \text{ Using } (ii) \right]\]
\[ = \frac{1}{x}\left( a + b \right)\]
\[ = \frac{2}{\left( a + b \right)}\left( a + b \right) \left[ \text{ Using } (i) \right]\]
\[ = 2\]
\[\]
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