A) \[x=y=z\]
B) \[x=y=-z\]
C) \[x=1;y=2;z=3\]
D) \[x=2;y=4;z=6\]
E) \[x=2y=3z\]
Correct Answer: A
Solution :
\[2{{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}}\]z Þ\[{{\tan }^{-1}}\left( \frac{2y}{1-{{y}^{2}}} \right)={{\tan }^{-1}}\left( \frac{x+z}{1-xz} \right)\] Þ\[\frac{2y}{1-{{y}^{2}}}=\frac{x+z}{1-xz}\] But \[2y=x+z\] \\[1-{{y}^{2}}=1-xz\]Þ \[{{y}^{2}}=xz\] \[\because xyz\] are both in G.P. and A.P., \ \[x=y=z\].
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