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If $a,b,c$ are respectively the ${p^{th}},{q^{th}}{r^{th}}$terms of an $A.P.,$ the $\left| {\,\begin{array}{*{20}{c}}a&p&1\\b&q&1\\c&r&1\end{array}\,} \right| = $
$1$
$-1$
$0$
$pqr$
Solution
(c) Let first term =$ A $ and common difference =$ D$
$\therefore a = A + (p – 1)D$, $b = A + (q – 1)D$, $c = A + (r – 1)D$
$\left| {\begin{array}{*{20}{c}}{\,a\,\,\,}&{p\,\,\,}&{1\,}\\{\,b\,\,\,}&{q\,\,\,}&1\\{\,c\,\,\,}&{r\,\,\,}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{\,A + (p – 1)D\,\,\,}&{p\,\,\,}&{1\,}\\{A + (q – 1)D\,\,\,}&{q\,\,\,}&1\\{A + (r – 1)D\,\,\,}&{r\,\,\,}&1\end{array}} \right|$
Operate ${C_1} \to {C_1} – D{C_2} + D{C_3}$
$ = \,\left| {\begin{array}{*{20}{c}}{\,A\,\,}&{p\,\,}&{1\,}\\{\,A\,\,}&{q\,\,}&1\\{\,A\,\,}&{r\,\,}&1\end{array}} \right| = A\left| {\begin{array}{*{20}{c}}{\,\,1}&{\,\,p}&{\,\,1\,}\\{\,1}&q&1\\{\,1}&r&1\end{array}} \right| = 0$.
