If in a triangle PQR, sinP, sinQ, sinR  are in A.P., then

From sine rule
$\frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}=K\text{ (say) }$
$\text{ Also }\left(\frac{1}{2}\right)a{P}_1=\mathrm{\Delta }\left(\text{ using }\frac{1}{2}\times \text{ base }\times \text{ height }=\mathrm{\Delta }\right)$
 

$\begin{array}{l}{P}_1-\frac{2\mathrm{\Delta }}{a}-\frac{2\mathrm{\Delta }}{K\sin P}\\ similarly ,P_2=\frac{2\mathrm{\Delta }}{a}=\frac{2\mathrm{\Delta }}{K\sin Q}\text{ and }{P}_3=\frac{2\mathrm{\Delta }}{K\sin R}\end{array}$
Since sin P, sinQ and sinR are in A.P., Hence P₁, P₂, P₃ will be in H.P.