Question

If the non-zero numbers a, b, c are in A.P. and ${\tan }^{-1}a,{\tan }^{-1}b,{\tan }^{-1}c$ are also in A.P., then prove that $a=b=c$ and $b^2=ac$.

Answer 1

$2b=a+c$ .$\left(1\right)$

and $2{\tan }^{-1}b={\tan }^{-1}a+{\tan }^{-1}c$

or $\frac{2b}{1-b^2}=\frac{a+c}{1-ac}$ Put $2b=a+c$

$\Rightarrow b^2=ac$ by $\left(1\right)$ i.e., a, b, c are in G.P.

Also $\Rightarrow 4b^2=4ac$

or $(a+c{)}^2-4ac=0$, by $\left(1\right)$

$(a-c{)}^2=0$

$\therefore a=c=b$, by $\left(1\right)$.