Sequences & Series
Arithmetic Progression - Word Problems
Grade 11
Question:
<p>In a cricket tournament 16 school teams participated. A sum of ₹8000 is to be awarded among themselves as prize money. If the last placed team is awarded ₹275 in prize money and the award increases by the same amount for successive finishing places, what amount will the first place team receive?</p>
<p>(a) ₹720</p>
<p>(b) ₹725</p>
<p>(c) ₹735</p>
<p>(d) ₹780</p>
Step-by-Step Solution
Key Concept: This is an arithmetic progression problem where the total sum of all prizes is fixed at ₹8000, the last place (16th position) gets ₹275, and the prize increases by a constant amount d for each successive finishing place. We need to find the prize for first place using the sum formula for AP.
<p><strong>Step 1:</strong> Identify the given information.</p><p>• Number of teams (n) = 16</p><p>• Total prize money = ₹8000</p><p>• Last placed team (16th position) receives a₁₆ = ₹275</p><p>• Prize increases by constant amount d for successive finishing places</p><p>• Need to find: Prize for first place (a₁)</p><p><strong>Step 2:</strong> Set up the arithmetic progression.</p><p>Since prizes increase for successive finishing places (moving from 16th to 1st), we have:</p><p>a₁₆ = 275 (last place)</p><p>a₁ = ? (first place)</p><p>Common difference = d (positive, since prizes increase)</p><p><strong>Step 3:</strong> Use the sum formula for arithmetic progression.</p><p>Sum of AP: S_n = (n/2)(first term + last term)</p><p>8000 = (16/2)(a₁ + 275)</p><p>8000 = 8(a₁ + 275)</p><p><strong>Step 4:</strong> Solve for a₁.</p><p>8000 = 8a₁ + 2200</p><p>8000 - 2200 = 8a₁</p><p>5800 = 8a₁</p><p>a₁ = 5800/8</p><p>a₁ = 725</p><p><strong>Step 5:</strong> Verification.</p><p>Average prize = 8000/16 = ₹500</p><p>Middle value (average of 1st and last) = (725 + 275)/2 = 1000/2 = 500 ✓</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b