# If a,b,c are the p^th. q^th, r^th terms in H.P. then prove that (q -r)bc+(r -p)ca+(p-q)ab=0

Solution: Let A and D be the first term and common difference of the corresponding AP. Now a ,b , c are respectively the pth , qth , and rth terms of HP.

$\\\therefore$ 1/a , 1/b , 1/c will be respectivvely the pth , qth  and rth terms of the corresponding AP.

$\\ \Rightarrow 1/a=A+(p-1)D .........(1)\\ \\ \Rightarrow 1/b=A+(q-1)D.......(2)\\ \\\Rightarrow 1/c=A+(r-1)D........(3)$

Subtracting Equation (3) from Equation(2) , we get

$\\ (1/b )-(1/c)=(q-r)D\Rightarrow bc(q-r)=(c-b)/D=-(b-c)/D$

LHS $\\=(q-r)bc+(r-p)ca+(p-q)ab=0$

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