Question

If the $pth,qth$ and $rth$ terms of an $H.P.$ be $a,b,c$ respectively, then value of
$\Delta =$ $\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|=0$

Answer 1

Given,

$\frac{1}{a}=\frac{1}{A}+(p-1)D$

$\frac{1}{b}=\frac{1}{A}+(q-1)D$

$\frac{1}{c}=\frac{1}{A}+(r-1)D$

$\Delta =\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|$

$=\left|\begin{array}{ccc}(b-a)c& (c-b)a& ab\\ p-q& q-r& r\\ 0& 0& 1\end{array}\right|$

$=(b-a)c(q-r)-(p-q)(c-b)a$

$\Delta =0$ (proved)