If in a $\triangle P Q R, \sin P ; \sin Q, \sin R$ are in $A P$, then

• A

  the altitudes are in AP
• Correct answer
  B

  the altitudes are in HP
• C

  the medians are in GP
• D

  the medians are in AP

Solution

By sine rule
$\begin{array}{ll} & \frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}=k \\ \text { Also, } & \frac{1}{2} a p_1=\Delta \\ \Rightarrow & \frac{2 \Delta}{a}=p_1 \\ \Rightarrow & p_1=\frac{2 \Delta}{k \sin P} \end{array}$ Similarly, $p_1=\frac{2 \Delta}{k \sin Q}$ and $p_3=\frac{2 \Delta}{k \sin R}$ Since, $\sin P, \sin Q, \sin R$ are in AP. $\therefore \quad \frac{1}{\sin P} \cdot \frac{1}{\sin Q}, \frac{1}{\sin R}$ are in $H P$ $\Rightarrow \frac{2 \Delta}{k \sin P}, \frac{2 \Delta}{k \sin Q}, \frac{2 \Delta}{k \sin R}$ are in HP $\Rightarrow p_1, p_2, p_3$ are in HP.

Hence, the correct answer is option: 'B'.