If $p,q,r$ are in GP and $ an^{- 1}p, an^{- 1}q, an^{- 1}r$ are in AP, then $p,q,r$ satisfies the relation

Question

If $p,q,r$ are in GP and $	an^{- 1}p,	an^{- 1}q,	an^{- 1}r$ are in AP, then $p,q,r$ satisfies the relation

Dr. Abubacker · Accepted Answer

(a)
Since, 2$	an^{- 1}{q = 	an^{- 1}{p + 	an^{- 1}r}}$
$\Rightarrow 	an^{- 1}\frac{2q}{1 - q^{2}} = 	an^{- 1}\frac{p + r}{1 - p^{r}}$
$\Rightarrow \ \ 2q = p + r$ [$	herefore q^{2} = pr$]
$\Rightarrow \ \ \ \ p,q,r\ are\ in$ AP.
But p, q, r are in GP.
$\Rightarrow p = q = r$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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