<p>If \(a_1, a_2, a_3, \ldots, a_n, \ldots\) are in GP, then the value of the determinant \[\begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{vmatrix}\] is</p>
Step-by-Step Solution
Key Concept: When terms are in GP with common ratio r, their logarithms form an AP with common difference log r. This transforms the determinant into one with linearly dependent rows, making it zero.
<p><strong>Step 1:</strong> Since a₁, a₂, a₃, ... are in GP with first term a₁ and common ratio r, we have aₙ = a₁·r^(n-1).</p><p><strong>Step 2:</strong> Taking logarithm: log aₙ = log a₁ + (n-1)log r. So the sequence {log aₙ} forms an AP with first term log a₁ and common difference d = log r.</p><p><strong>Step 3:</strong> Denote log aₙ = α. Then:<br>• Row 1: α, α+d, α+2d<br>• Row 2: α+3d, α+4d, α+5d<br>• Row 3: α+6d, α+7d, α+8d</p><p><strong>Step 4:</strong> We can verify that Row 2 = Row 1 + (3d)·[1,1,1] and Row 3 = Row 2 + (3d)·[1,1,1]. More directly, Row 3 - 2·Row 2 + Row 1 = [0,0,0], showing the rows are linearly dependent.</p><p><strong>Step 5:</strong> Since the three rows are linearly dependent, the determinant = 0.</p><p>∴ Answer: A</p>
Correct Answer: A