If left| begin{array}{ccc} a+1 & a+2 & a+p a | vedclass

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(A) Given the determinant $\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \ a+2 & a+3 & a+q \ a+3 & a+4 & a+r \end{array} ight| = 0$. Apply ro

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$AP$

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(A) Given the determinant $\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \ a+2 & a+3 & a+q \ a+3 & a+4 & a+r \end{array} ight| = 0$. Apply row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$: $\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \ 1 & 1 & q-p \ 1 & 1 & r-q \end{array} ight| = 0$. Since two rows ($R_2$ and $R_3$) have identical elements in the first two columns,we can perform $R_3 ightarrow R_3 - R_2$: $\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \ 1 & 1 & q-p \ 0 & 0 & (r-q) - (q-p) \end{array} ight| = 0$. Expanding along the third row: $0 - 0 + (r - q - q + p) 	imes ((a+1)(1) - (a+2)(1)) = 0$. $(r + p - 2q) 	imes (-1) = 0$. This implies $r + p - 2q = 0$,or $p + r = 2q$. This is the condition for $p, q, r$ to be in $AP$.

If $\left| \begin{array}{ccc} a+1 & a+2 & a+p \\ a+2 & a+3 & a+q \\ a+3 & a+4 & a+r \end{array} \right| = 0$,then $p, q, r$ are in :

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