Let $AB$ be divided internally and externally at $R$ and $S$ in the same ratio. Then $AR,AB,AS$ are in

$\overline{AB}$ is divided  by $R\&S$ internally & Externally in the same ratio

Let $R,S$ divide $\overline{AB}$ harmonically in the ratio $m:n$

$\Rightarrow \frac{AR}{RB}=\frac{m}{n}=\frac{AS}{SB}\Rightarrow \frac{RB}{AR}=\frac{n}{m}=\frac{SB}{AS}$

$\Rightarrow \frac{RB}{AR}+1=\frac{n}{m}+1\Rightarrow \frac{RB+AR}{AR}=\frac{n+m}{m}$

$1-\frac{SB}{AS}=1-\frac{n}{m}\frac{AS-SB}{AS}=\frac{m-n}{m}\Rightarrow \frac{AB}{AR}=\frac{m+n}{m},\frac{AB}{AS}=\frac{m-n}{m}$

$\frac{AB}{AR}+\frac{AB}{AS}=\frac{2m}{m}\Rightarrow \frac{1}{AR}+\frac{1}{AS}=\frac{2}{AB}$

$\Rightarrow AR,AB,AS$  are in H.P.