<p>Between two numbers whose sum is <span>\(2\dfrac{1}{6}\)</span>, an even number of arithmetic means are inserted. The sum of these means exceeds their number by unity. How many means are there?</p>
Step-by-Step Solution
Key Concept: If n arithmetic means are inserted between two numbers a and b, the sum of means equals n times the middle value of the AP formed. Set up an equation using: sum of means = (n/2)(a+b) and apply the condition that sum exceeds count by 1.
<p><strong>Step 1:</strong> Let the two numbers be a and b with a + b = 2 + 1/6 = 13/6. Let n arithmetic means be inserted (n is even).</p><p><strong>Step 2:</strong> The n means form an AP with first term (a + d) and last term (b - d), where d is the common difference of the complete sequence. The complete AP has (n+2) terms.</p><p><strong>Step 3:</strong> Sum of all (n+2) terms = (n+2)/2 × (a+b) = (n+2)/2 × 13/6</p><p><strong>Step 4:</strong> Sum of n means alone = Total sum - a - b = (n+2)/2 × 13/6 - 13/6 = 13/6 × [(n+2)/2 - 1] = 13/6 × n/2 = 13n/12</p><p><strong>Step 5:</strong> Given condition: Sum of means = Number of means + 1</p><p>13n/12 = n + 1</p><p><strong>Step 6:</strong> Solving: 13n/12 - n = 1 → n/12 = 1 → n = 12</p><p><strong>Step 7:</strong> Verify: n = 12 is even ✓, and 13(12)/12 = 13 = 12 + 1 ✓</p><p>∴ Answer: <strong>12</strong></p>
Correct Answer: 12