<p><em>l</em>, <em>m</em> and <em>n</em> are the <em>p</em>th, <em>q</em>th and <em>r</em>th terms of an GP and all positive, then \(\begin{vmatrix} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1 \end{vmatrix}\) equals</p>
Step-by-Step Solution
Key Concept: Since l, m, n are terms of a GP, they satisfy l = a·b^(p-1), m = a·b^(q-1), n = a·b^(r-1). Taking logarithms converts the geometric progression property into a linear relationship: log l = log a + (p-1)log b, which means the rows of the determinant become linearly dependent.
<p><strong>Step 1:</strong> Express GP terms using the general form.</p><p>If l, m, n are pth, qth, rth terms of a GP with first term a and common ratio b, then:</p><p>l = a·b^(p-1), m = a·b^(q-1), n = a·b^(r-1)</p><p><strong>Step 2:</strong> Take logarithms of all three terms.</p><p>log l = log a + (p-1)log b</p><p>log m = log a + (q-1)log b</p><p>log n = log a + (r-1)log b</p><p><strong>Step 3:</strong> Observe the linear relationship.</p><p>Notice that if we write log l = log a - log b + p·log b, we can see that:</p><p>log l = (log a - log b) + p·(log b)</p><p>log m = (log a - log b) + q·(log b)</p><p>log n = (log a - log b) + r·(log b)</p><p><strong>Step 4:</strong> Recognize linear dependence in the determinant.</p><p>The first column can be expressed as a linear combination of the second and third columns. Specifically, the three rows satisfy the relation:</p><p>R₁: [log l, p, 1], R₂: [log m, q, 1], R₃: [log n, r, 1]</p><p>Since log l = (log a - log b)·1 + (log b)·p, the rows are linearly dependent.</p><p><strong>Step 5:</strong> Apply the determinant property.</p><p>When rows (or columns) of a determinant are linearly dependent, the determinant equals zero.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: D