<p>If \(a\), \(b\), \(c\) are positive and are the \(p\)th, \(q\)th, and \(r\)th terms, respectively, of a G.P., then \(\Delta = \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix}\) is</p>
Step-by-Step Solution
Key Concept: In a G.P., the pth, qth, and rth terms satisfy the relationship: log(term) = log(first term) + (index-1)·log(common ratio), which makes the first column a linear combination of the second and third columns, rendering them linearly dependent.
<p><strong>Step 1:</strong> Since a, b, c are pth, qth, rth terms of a G.P., let the first term be A and common ratio be d.</p><p>Then: a = A·d^(p-1), b = A·d^(q-1), c = A·d^(r-1)</p><p><strong>Step 2:</strong> Taking logarithms:</p><p>log a = log A + (p-1)log d</p><p>log b = log A + (q-1)log d</p><p>log c = log A + (r-1)log d</p><p><strong>Step 3:</strong> Rewrite the determinant columns:</p><p>Column 1 = (log A)·Column 3 + (log d)·(Column 2 - Column 3)</p><p>This shows Column 1 is a linear combination of Columns 2 and 3.</p><p><strong>Step 4:</strong> Alternatively, perform row operations: R₂ - R₁ and R₃ - R₁:</p><p>The new rows become:</p><p>R₂: [log(b/a), q-p, 0] = [(q-p)log d, q-p, 0]</p><p>R₃: [log(c/a), r-p, 0] = [(r-p)log d, r-p, 0]</p><p><strong>Step 5:</strong> Since two rows are now linearly dependent (both have 0 in the third column and the first two entries are proportional), the determinant equals 0.</p><p>∴ Answer: <strong>A (Δ = 0)</strong></p>
Correct Answer: A