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