If pth, qth and rth terms of a HP be respectively`a,b` and `c`, has prove that `(q-r)bc+(r-p)ca+(p-q)ab=0`.

Explanation

Let A and D be the first term and common difference of the corrsponding AP. Now, `a,b,c` are respectively the pth, qth and rth terms of HP.
`therefore (1)/(a),(1)(b),(1)/(c )` will be respectively the pth, qth and rth terms of the corresponding AP.
`implies (1)/(a)=A+(p-1)D " " "........(i)"`
`(1)/(b)=A+(q-1)D " " ".......(ii)`
`(1)/(c)=A+(r-1)D " " "......(iii)`
On sutracting Eq. (iii) from Eq.(ii), we get
`(1)/(b)-(1)/(c)=(q-r)D implies bc (q-r)=((c-b))/(D)=-((b-c))/(D)`
So, `LHS=(q-r)bc +(r-p)ca+(p-q)ab` ltbegt `=-(1)/(D){b-c+c-a+a-b}=0=RHS`