If a, b, c are in A.P., then the value of the determinant \(\begin{vmatrix} x + 2 & x + 3 & x + 2a \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c \end{vmatrix}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Applying \(R_1 \to R_1 + R_3 - 2R_2\), which does not change the value of the determinant, the new elements of \(R_1\) are
\((x + 2) + (x + 4) - 2(x + 3) = 0\)
\((x + 3) + (x + 5) - 2(x + 4) = 0\)
\((x + 2a) + (x + 2c) - 2(x + 2b) = 2(a + c - 2b)\)
Since a, b, c are in A.P., \(2b = a + c\), so the third element is also 0.
\( \displaystyle \Delta = \begin{vmatrix} 0 & 0 & 0 \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c \end{vmatrix} \)
A determinant with a row of zeros is zero, so \(\Delta = 0\) for every value of x.