Question

If $p^{th},q^{th},r^{th}$ terms of an HP. be $a,b,c$ respectively, prove that
$(q-r)bc+(r-p)ac+(p-q)ab=0$

Answer 1

Let ${}^{\prime }{x}^{\prime }$ be the first term and ${}^{\prime }{d}^{\prime }$ be the common difference of the corresponding A.P..
so $\frac{1}{a}=x+(p-1)d$ (i)
$\frac{1}{b}=x+(q-1)d$ (ii)
$\frac{1}{c}=x+(r-1)d$ (iii)
(i)-(ii) $\Rightarrow$ $ab(p-q)d=b-a$ (iv)
(ii)-(iii) $\Rightarrow$ $bc(q-r)d=c-b$ (v)
(iii)-(i) $\Rightarrow$ $ac(r-p)d=a-c$ (vi)
(iv)+(v)+(vi) gives
$bc(q-r)+ac(r-p)+ab(p-q)=0.$