A) \[p=q=r\]
B) \[p\ne q\ne r\]
C) \[p+q=r\]
D) None of these
Correct Answer: A
Solution :
\[p,q,r\in \text{G}\text{.}\,\text{P}\text{.}\], \[\therefore {{q}^{2}}=pr\] Also \[{{\tan }^{-1}}p,{{\tan }^{-1}}q,\] \[{{\tan }^{-1}}r\in \]A.P. Þ \[{{\tan }^{-1}}p+{{\tan }^{-1}}r=2{{\tan }^{-1}}q\] Þ \[p+r=2q\Rightarrow p,q,r\] are in A.P. Now p, q, r are both in A.P and G.P., which is possible only, if \[p=q=r\].
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