Question

If p,q,r are in GP and \[{\tan ^{ - 1}}p,\,{\tan ^{ - 1}}q,{\tan ^{ - 1}}r\] are in AP, then p,q,r satisfies the relation,
A. \[p = q = r\]
B. \[p \ne q \ne r\]
C. \[2q = p + r\]
D. None of these

Answer 1

Hint: Here the given question express the condition for three variables, and need to find the relation between them, here we know the formulae for common difference of AP and common ratio of GP, here we can form the equation from there and then after rearranging the values we can reach to the solution.
Formulae Used:
\[
   \Rightarrow for\,p,q,r\,in\,AP \\
   \Rightarrow common\,difference \to 2q = p + r \\
 \]
\[
   \Rightarrow for\,p,q,r\,in\,GP \\
   \Rightarrow common\,ratio \to \dfrac{q}{p} = \dfrac{r}{q} \\
 \]
\[ \Rightarrow 2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\]
\[ \Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]

Complete step by step answer:
Here the given question says that three variable, are in GP and there inverse in “tan” function is in AP, so to find the relation between the variables we need to write the common difference of AP and then write the common ratio of GP, and then solve further, on solving we get:
Common ratio of GP:
\[
   \Rightarrow \dfrac{q}{p} = \dfrac{r}{q} \\
   \Rightarrow {q^2} = p \times r \\
 \]
Common difference of AP:
\[
   \Rightarrow {\tan ^{ - 1}}q - {\tan ^{ - 1}}p = {\tan ^{ - 1}}r - {\tan ^{ - 1}}q \\
   \Rightarrow 2{\tan ^{ - 1}}q = {\tan ^{ - 1}}p + {\tan ^{ - 1}}r \\
 \]
Now using the formulae of “tan” function here we get:
\[
   \Rightarrow 2{\tan ^{ - 1}}q = {\tan ^{ - 1}}p + {\tan ^{ - 1}}r \\
   \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2q}}{{1 - {q^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{p + r}}{{1 - pr}}} \right) \\
 \]
Now putting values from the common ratio of GP, we get:
\[
   \Rightarrow we\,have,\,{q^2} = p \times r \\
   \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2q}}{{1 - pr}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{p + r}}{{1 - pr}}} \right) \\
 \]
Since both side the function is inverse of “tan” hence both the angles will be same, equating then we get:
\[
   \Rightarrow \left( {\dfrac{{2q}}{{1 - pr}}} \right) = \left( {\dfrac{{p + r}}{{1 - pr}}} \right) \\
   \Rightarrow 2q = p + r \\
 \]
Here we get the answer for the given question, the relation obtained is the same as that of common difference in AP.

So, the correct answer is “Option C”.

Note: Here in the given question we need to solve two equations simultaneously, in such a type of question, rearranging of values is the best possible option of solving the question, here we put the values from the first equation into the second and reach to the final solution.