<p>Let \(\{a_n\}\) \((n \geq 1)\) be a sequence such that \(a_1 = 1\), and \(3a_{n+1} - 3a_n = 1\) for all \(n \geq 1\). Then find the value of \(a_{2002}\).</p>
Step-by-Step Solution
Key Concept: Recognize this as an arithmetic progression with common difference d = 1/3 derived from the recurrence relation 3a_{n+1} - 3a_n = 1, which simplifies to a_{n+1} - a_n = 1/3.
<p><strong>Step 1:</strong> Simplify the recurrence relation.</p><p>Given: 3a_{n+1} - 3a_n = 1</p><p>Dividing both sides by 3: a_{n+1} - a_n = 1/3</p><p>This shows {a_n} is an arithmetic sequence with common difference d = 1/3.</p><p><strong>Step 2:</strong> Apply the arithmetic sequence formula.</p><p>For an arithmetic sequence: a_n = a_1 + (n-1)d</p><p>With a_1 = 1 and d = 1/3:</p><p>a_{2002} = 1 + (2002 - 1) × (1/3)</p><p>a_{2002} = 1 + 2001 × (1/3)</p><p>a_{2002} = 1 + 667</p><p>a_{2002} = 668</p><p><strong>∴ Answer: 668</strong></p>
Correct Answer: 668