Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let \(T_r\) be the \(r\)th term of an AP whose first term is \(a\) and common difference is \(d\). If for some positive integers \(m, n, m \neq n\), \(T_m = \dfrac{1}{n}\) and \(T_n = \dfrac{1}{m}\), then \(a - d\) equals</p>
<p>0</p>
<p>1</p>
<p>\(\dfrac{1}{mn}\)</p>
<p>\(\dfrac{1}{m} + \dfrac{1}{n}\)</p>
Step-by-Step Solution
Key Concept: Use the AP term formula T_r = a + (r-1)d to set up two equations from the given conditions, then subtract them to find a relationship between a and d that directly yields a - d.
<p><strong>Step 1:</strong> Write the general term formula. For AP with first term a and common difference d:</p><p>T_r = a + (r-1)d</p><p><strong>Step 2:</strong> Use the given conditions:</p><p>T_m = a + (m-1)d = 1/n ... (1)</p><p>T_n = a + (n-1)d = 1/m ... (2)</p><p><strong>Step 3:</strong> Subtract equation (2) from equation (1):</p><p>[a + (m-1)d] - [a + (n-1)d] = 1/n - 1/m</p><p>(m-1)d - (n-1)d = 1/n - 1/m</p><p>(m - n)d = (m - n)/(mn)</p><p><strong>Step 4:</strong> Since m ≠ n, divide both sides by (m - n):</p><p>d = 1/(mn)</p><p><strong>Step 5:</strong> Add equations (1) and (2):</p><p>2a + (m+n-2)d = 1/n + 1/m = (m+n)/(mn)</p><p><strong>Step 6:</strong> Substitute d = 1/(mn):</p><p>2a + (m+n-2)·(1/mn) = (m+n)/(mn)</p><p>2a = (m+n)/(mn) - (m+n-2)/(mn) = 2/(mn)</p><p>a = 1/(mn)</p><p><strong>Step 7:</strong> Calculate a - d:</p><p>a - d = 1/(mn) - 1/(mn) = 0</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: A