Question

If \(p, q, r\) are in A.P., then the value of \[ \begin{vmatrix} x+4 & x+9 & x+p \\ x+5 & x+10 & x+q \\ x+6 & x+11 & x+r \end{vmatrix} \] is:

• A. \(x + 15\)
• B. \(x + 20\)
• C. \(x + p + q + r\)
• D. None of these

Hint:

In determinant problems involving A.P., try using column/row operations to exploit arithmetic progression relations — often, it reduces to a zero determinant.

Answer 1

The Correct Option is D

Given that \(p, q, r\) are in A.P., we have \(q = \frac{p + r}{2}\).
Let’s simplify the determinant:
First, notice that in each column, the \(x\) term appears in every element. We can factor \(x\) out as part of column operations, but here it appears in all terms, so it will be eliminated when we take differences.
Perform \(C_1 \to C_1 - C_2\) and \(C_3 \to C_3 - C_2\):
\[ \begin{vmatrix} -5 & x+9 & p - 9 \\ -5 & x+10 & q - 10 \\ -5 & x+11 & r - 11 \end{vmatrix} \] Now, \(C_1\) has constant entries \((-5, -5, -5)\), making it proportional, which means the determinant will simplify using cofactor expansion.
Also, using the property of determinants, if two columns are linearly dependent, the determinant will be zero.
Since \(p, q, r\) are in A.P., \(q - 10 = \frac{(p - 9) + (r - 11)}{2}\), which confirms a relation between entries in \(C_3\) and \(C_1, C_2\).
After simplification, the determinant evaluates to \(0\), which means it does not match any form \(x + \text{constant}\) or \(x + p + q + r\).
Thus, the answer is (D) None of these.