Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let \(a_1, a_2, \ldots, a_{30}\) be an A.P., \(S = \displaystyle\sum_{i=1}^{30} a_i\) and \(T = \displaystyle\sum_{i=1}^{15} a_{(2i-1)}\). If \(a_5 = 27\) and \(S - 2T = 75\), then \(a_{10}\) is equal to __________.</p>
Step-by-Step Solution
Key Concept: Recognize that T is the sum of odd-positioned terms, so S - 2T isolates the sum of even-positioned terms. Use this to create a system of equations with the A.P. properties (first term and common difference).
<p><strong>Step 1:</strong> Set up A.P. with first term <em>a</em> and common difference <em>d</em>. Then aᵢ = a + (i-1)d.</p><p><strong>Step 2:</strong> Express S = Σ(i=1 to 30) aᵢ = 30a + d(0+1+2+...+29) = 30a + 435d</p><p><strong>Step 3:</strong> Express T = Σ(i=1 to 15) a₍₂ᵢ₋₁₎ = a₁ + a₃ + a₅ + ... + a₂₉ (odd-positioned terms, 15 terms)<br/>These are: a, a+2d, a+4d, ..., a+28d<br/>T = 15a + d(0+2+4+...+28) = 15a + 210d</p><p><strong>Step 4:</strong> Calculate S - 2T = 30a + 435d - 2(15a + 210d) = 30a + 435d - 30a - 420d = 15d</p><p><strong>Step 5:</strong> From S - 2T = 75, we get 15d = 75 → <strong>d = 5</strong></p><p><strong>Step 6:</strong> Use a₅ = 27. Since a₅ = a + 4d, we have: a + 4(5) = 27 → a + 20 = 27 → <strong>a = 7</strong></p><p><strong>Step 7:</strong> Find a₁₀ = a + 9d = 7 + 9(5) = 7 + 45 = <strong>52</strong></p><p>∴ Answer: <strong>52</strong></p>
Correct Answer: 52