If a,b,c are the pth,qth and rth terms of an H.P, then the lines bcx+py+1=0, cax+qy+1=0 and abx+ry+1=0,

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If a,b,c are the pth,qth and rth terms of an H.P, then the lines bcx+py+1=0, cax+qy+1=0 and abx+ry+1=0,

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Given a,b,c are pth,qth,rth terms of H.P. respectively.⇒a1​,b1​,c1​ are pth,qth,rth terms of A.P. respectively.Let A be the first term and d be the common differenceTp​=A+(p−1)da1​=A+(p−1)d         .....(1)Tq​=A+(q−1)db1​=A+(q−1)d         .....(2)Tr​=A+(r−1)dc1​=A+(r−1)d         .....(3)Subtracting (2) from (1),  we geta1​−b1​=(p−q)d⇒ab(p−q)=db−a​       .....(4)Subtracting (3) from (2),  we getb1​−c1​=(q−r)d⇒bc(q−r)=dc−b​      .....(5)Subtracting (3) from (1),  we geta1​−c1​=(p−r)d⇒ac(p−r)=dc−a​      .....(6)Now, consider ∣∣​bccaab​pqr​111​∣∣​(The above matrix is the matrix representation of the given equations)Expanding along first column, we get=bc(q−r)−ca(p−r)+ab(p−q)=dc−b​−dc−a​+db−a​=0Hence, the given lines are concurrent.Hence, option A.

Question Text	If $a, b, c$ are the $p^{t h}, q^{t h}$ and $r^{t h}$ terms of an $H . P$ , then the lines $b c x + p y + 1 = 0, c a x + q y + 1 = 0$ and $ab x + ry + 1 = 0$ ,
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If $a, b, c$ are the $p^{t h}, q^{t h}$ and $r^{t h}$ terms of an $H . P$ , then the lines $b c x + p y + 1 = 0, c a x + q y + 1 = 0$ and $ab x + ry + 1 = 0$ ,

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