A) the altitudes are in\[AP\]
B) the altitudes are in\[HP\]
C) the medians are in\[GP\]
D) the medians are in\[AP\]
Correct Answer: B
Solution :
By sine rule \[\frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}=k\] (say) Also, \[\frac{1}{2}a{{p}_{1}}=\Delta \] \[\Rightarrow \] \[\frac{2\Delta }{a}={{p}_{1}}\] \[\Rightarrow \] \[{{p}_{1}}=\frac{2\Delta }{k\sin P}\] 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\[AP\]. \[\therefore \] \[\frac{1}{\sin P}\cdot \frac{1}{\sin Q}\cdot \frac{1}{\sin R}\]are in\[HP\] \[\Rightarrow \] \[\frac{2\Delta }{k\sin P},\,\,\frac{2\Delta }{k\sin Q},\,\,\frac{2\Delta }{k\sin R}\]are in\[HP\] \[\Rightarrow \,\,{{p}_{1}},\,{{p}_{2}},\,{{p}_{3}}\] are in HP.You need to login to perform this action.
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