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(B) Let $\Delta$ be the area of $\Delta ABC$ and $p_1, p_2, p_3$ be the altitudes from $A, B, C$ respectively.We know that $\Delta = \frac{1}{2} p_1 a

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(B) Let $\Delta$ be the area of $\Delta ABC$ and $p_1, p_2, p_3$ be the altitudes from $A, B, C$ respectively.We know that $\Delta = \frac{1}{2} p_1 a = \frac{1}{2} p_2 b = \frac{1}{2} p_3 c$.Therefore,$p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.Given that $p_1, p_2, p_3$ are in harmonic progression $(H.P.)$,$\frac{1}{p_1}, \frac{1}{p_2}, \frac{1}{p_3}$ are in arithmetic progression $(A.P.)$.Substituting the values,$\frac{a}{2\Delta}, \frac{b}{2\Delta}, \frac{c}{2\Delta}$ are in $A.P.$This implies $a, b, c$ are in $A.P.$Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.Thus,$2R \sin A, 2R \sin B, 2R \sin C$ are in $A.P.$Therefore,$\sin A, \sin B, \sin C$ are in arithmetic progression.

In $\Delta ABC$,if the altitudes from $A, B, C$ to the opposite sides are in harmonic progression,then $\sin A, \sin B, \sin C$ are in ............. progression.

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