Sequence And Series Question 479
Question: If $ x,y,z $ are in A.P. and $ {{\tan }^{-1}}x,{{\tan }^{-1}}y $ and $ {{\tan }^{-1}}z $ are also in A.P., then
[Kerala (Engg.) 2005]
Options:
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 \quad y=3z $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2{{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}} $ z
Þ $ {{\tan }^{-1}}( \frac{2y}{1-y^{2}} )={{\tan }^{-1}}( \frac{x+z}{1-xz} ) $
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 x,y,z $ are both in G.P. and A.P., \ $ x=y=z $ .
BETA
