Question

If \( a_1, a_2, a_3, \dots \) are in A.P., then the value of \( \begin{vmatrix} a_1 & a_2 & 1 \\ a_2 & a_3 & 1 \\ a_3 & a_4 & 1 \end{vmatrix} \) is equal to:

• A. \( a_4 - a_1 \)
• B. \( \frac{a_1 + a_4}{2} \)
• C. \( 1 \)
• D. \( \frac{a_2 + a_3}{2} \)
• E. \( 0 \)

Hint:

Determinants involving A.P. terms often resolve to zero because the "constant increase" creates linear dependency between rows.

Answer 1

Answer: (E)

Concept: For an Arithmetic Progression (A.P.), the difference between consecutive terms is constant: \( a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d \). We can use row operations to simplify the determinant.

Step 1: Perform row operations.
Apply \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_2 \): \[ \begin{vmatrix} a_1 & a_2 & 1 a_2 - a_1 & a_3 - a_2 & 1 - 1 a_3 - a_2 & a_4 - a_3 & 1 - 1 \end{vmatrix} = \begin{vmatrix} a_1 & a_2 & 1 \\ d & d & 0 \\ d & d & 0 \end{vmatrix} \]

Step 2: Analyze the result.
The second row and the third row are now identical (\( d, d, 0 \)).

Step 3: Conclusion.
In linear algebra, if any two rows (or columns) of a determinant are identical or proportional, the value of the determinant is always zero. \[ \text{Determinant} = 0 \]