Sequences & Series
Arithmetic progression and means
Grade 11
Question:
<p><strong>897.</strong> Let \(A_r\), \(r = 1, 2, \ldots, 29\) be arithmetic means between 303 and \(-57\) where \(A_r > A_{r+1}\) \(\forall\) \(r = 1, 2, \ldots, 28\). If \(S\) be the sum of these means, then the value of \(\left[\dfrac{S}{(A_{14}-12)|A_r|_{\min}}\right]\).</p><p>[Note: \([k]\) denotes greatest integer less than or equal to \(k\) and \(|A_r|_{\min}\) denotes the minimum value of \(|A_r|\).]</p>
Step-by-Step Solution
Key Concept: Since 29 arithmetic means are inserted between 303 and -57 with decreasing order, they form an AP with 31 terms total. The sum of an AP depends only on first and last terms, and the minimum absolute value occurs at the term closest to zero.
<p><strong>Step 1:</strong> Set up the AP with 31 terms: 303, A₁, A₂, ..., A₂₉, -57 (decreasing order)</p><p><strong>Step 2:</strong> Find common difference: d = (-57 - 303)/(31 - 1) = -360/30 = -12</p><p><strong>Step 3:</strong> General term: Aᵣ = 303 + r(-12) = 303 - 12r</p><p><strong>Step 4:</strong> Calculate sum S of the 29 means:</p><p>S = A₁ + A₂ + ... + A₂₉ = Σ(303 - 12r) for r=1 to 29</p><p>S = 29(303) - 12(1+2+...+29) = 8787 - 12(435) = 8787 - 5220 = 3567</p><p><strong>Step 5:</strong> Find A₁₄: A₁₄ = 303 - 12(14) = 303 - 168 = 135</p><p>So A₁₄ - 12 = 135 - 12 = 123</p><p><strong>Step 6:</strong> Find |Aᵣ|_min (minimum absolute value):</p><p>Aᵣ = 303 - 12r = 0 when r = 303/12 = 25.25</p><p>Check r = 25: A₂₅ = 303 - 300 = 3</p><p>Check r = 26: A₂₆ = 303 - 312 = -9</p><p>Therefore |Aᵣ|_min = min(|3|, |-9|) = 3</p><p><strong>Step 7:</strong> Calculate the expression:</p><p>[S/((A₁₄ - 12)|Aᵣ|_min)] = [3567/(123 × 3)] = [3567/369] = [9.659...] = 9</p><p>∴ Answer: <strong>9</strong></p>
Correct Answer: 9