<p>If \(a_1, a_2, \ldots, a_{4001}\) are in arithmetic progression and \(\dfrac{1}{a_1 a_2} + \dfrac{1}{a_2 a_3} + \ldots + \dfrac{1}{a_{4000} a_{4001}} = 10\) and \(a_2 + a_{4000} = 50\). Find the value of \(|a_1 - a_{4001}|\).</p>
Step-by-Step Solution
Key Concept: Use the telescoping series property: when terms are in AP with common difference d, the sum 1/(a_i·a_{i+1}) telescopes as (1/d)·(1/a_i - 1/a_{i+1}). Also apply the property that in an AP with odd number of terms, the middle term equals the average of equidistant terms.
<p><strong>Step 1: Set up the AP.</strong> Let the common difference be d. Then a_n = a_1 + (n-1)d for n = 1, 2, ..., 4001.</p><p><strong>Step 2: Apply the telescoping series formula.</strong> For consecutive terms in AP with common difference d: $$\frac{1}{a_i a_{i+1}} = \frac{1}{d}\left(\frac{1}{a_i} - \frac{1}{a_{i+1}}\right)$$</p><p>Therefore: $$\sum_{i=1}^{4000} \frac{1}{a_i a_{i+1}} = \frac{1}{d}\left(\frac{1}{a_1} - \frac{1}{a_{4001}}\right) = \frac{1}{d} \cdot \frac{a_{4001} - a_1}{a_1 a_{4001}} = 10$$</p><p><strong>Step 3: Relate to the common difference.</strong> Since a_{4001} - a_1 = 4000d, we have: $$\frac{1}{d} \cdot \frac{4000d}{a_1 a_{4001}} = 10$$</p><p>$$\frac{4000}{a_1 a_{4001}} = 10$$</p><p>$$a_1 a_{4001} = 400$$</p><p><strong>Step 4: Use the constraint a_2 + a_{4000} = 50.</strong> In an AP with 4001 terms, the 2nd and 4000th terms are equidistant from the middle (2001st term). Note that a_2 + a_{4000} = a_1 + d + a_1 + 3999d = 2a_1 + 4000d = 50.</p><p>Also, by symmetry in AP: a_1 + a_{4001} = a_2 + a_{4000} = 50.</p><p><strong>Step 5: Solve for |a_1 - a_{4001}|.</strong> Let a_1 = x and a_{4001} = y. We have:</p><p>• x + y = 50</p><p>• xy = 400</p><p>These are roots of: t² - 50t + 400 = 0</p><p>Using the quadratic formula: $$t = \frac{50 ± \sqrt{2500 - 1600}}{2} = \frac{50 ± \sqrt{900}}{2} = \frac{50 ± 30}{2}$$</p><p>So a_1 = 40, a_{4001} = 10 (or vice versa).</p><p>$$|a_1 - a_{4001}| = |40 - 10| = 30$$</p><p><strong>∴ Answer: 30</strong></p>
Correct Answer: 30