If a, b, c are respectively the p^{th}, q^{th | vedclass

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(C) Let the first term be $A$ and the common difference be $D$.Then,$a = A + (p - 1)D$,$b = A + (q - 1)D$,and $c = A + (r - 1)D$.The given determinant

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$0$

Vedclass · Answer

(C) Let the first term be $A$ and the common difference be $D$.Then,$a = A + (p - 1)D$,$b = A + (q - 1)D$,and $c = A + (r - 1)D$.The given determinant is $\Delta = \left| \begin{array}{ccc} A + (p - 1)D & p & 1 \ A + (q - 1)D & q & 1 \ A + (r - 1)D & r & 1 \end{array} ight|$.Applying the column operation $C_1 	o C_1 - (A - D)C_3$,we get:$\Delta = \left| \begin{array}{ccc} pD & p & 1 \ qD & q & 1 \ rD & r & 1 \end{array} ight|$.Taking $D$ common from $C_1$,we get $\Delta = D \left| \begin{array}{ccc} p & p & 1 \ q & q & 1 \ r & r & 1 \end{array} ight|$.Since column $C_1$ and $C_2$ are identical,the value of the determinant is $0$.

If $a, b, c$ are respectively the $p^{th}, q^{th}, r^{th}$ terms of an $A.P.$,then $\left| \begin{array}{ccc} a & p & 1 \\ b & q & 1 \\ c & r & 1 \end{array} \right| = $

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