If p, q, r are in A.P., then the value of \(\begin{vmatrix}y + 1 & y + 2 & y + p\\ y + 2 & y + 3 & y + q\\ y + 3 & y + 4 & y + r\end{vmatrix}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Since p, q, r are in A.P., \(2q = p + r\).
Apply the row operation \(R_1 \to R_1 + R_3 - 2R_2\). The new first row is:
\((y + 1) + (y + 3) - 2(y + 2) = 0\)
\((y + 2) + (y + 4) - 2(y + 3) = 0\)
\((y + p) + (y + r) - 2(y + q) = p + r - 2q = 0\)
The first row is now all zeros, so the determinant is zero for every y.