MHT CET · Maths · Sequences and Series
If \(x, y, \mathrm{z}\) are in A.P. and \(\tan ^{-1} x, \tan ^{-1} y\) and \(\tan ^{-1} z\) are also in A.P., then
- A \(x=y=\mathrm{z}\)
- B \(2 x=3 y=6 z\)
- C \(6 x=3 y=2 z\)
- D \(6 x=4 y=3 z\)
Answer & Solution
Correct Answer
(A) \(x=y=\mathrm{z}\)
Step-by-step Solution
Detailed explanation
Given, \(x, y, \mathrm{z}\) are in A.P.
\(\therefore 2 y=x+z... (i)\)
Also,
\(\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z \text { are in A.P. } \)
\( \therefore 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} z \)
\( \Rightarrow \tan ^{-1}\left(\frac{2 y}{1-y^2}\right)=\tan ^{-1}\left(\frac{x+z}{1-x z}\right) \)
\( \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-x z} \)
\( \Rightarrow \frac{2 y}{1-y^2}=\frac{2 y}{1-x z}... [From (i)] \)
\( \Rightarrow 1-y^2=1-x z \)
\( \Rightarrow y^2=x z\)
\(\therefore x, y, \mathrm{z} \text { are in G.P. }... (ii)\)
From (i) and (ii),
we get \(x=y=\mathrm{z}\)
\(\therefore 2 y=x+z... (i)\)
Also,
\(\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z \text { are in A.P. } \)
\( \therefore 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} z \)
\( \Rightarrow \tan ^{-1}\left(\frac{2 y}{1-y^2}\right)=\tan ^{-1}\left(\frac{x+z}{1-x z}\right) \)
\( \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-x z} \)
\( \Rightarrow \frac{2 y}{1-y^2}=\frac{2 y}{1-x z}... [From (i)] \)
\( \Rightarrow 1-y^2=1-x z \)
\( \Rightarrow y^2=x z\)
\(\therefore x, y, \mathrm{z} \text { are in G.P. }... (ii)\)
From (i) and (ii),
we get \(x=y=\mathrm{z}\)
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