<p>If <i>n</i> arithmetic means are inserted between 1 and 31 such that the 7th mean : the <span>\((n-1)\)</span>th mean = 5 : 9, then find <i>n</i>.</p>
Step-by-Step Solution
Key Concept: When n arithmetic means are inserted between two numbers, they form an AP with (n+2) terms total. Use the ratio condition to set up an equation relating the 7th and (n-1)th means to find n.
<p><strong>Step 1:</strong> Set up the AP. When n means are inserted between 1 and 31, we get an AP with (n+2) terms: 1, A₁, A₂, ..., Aₙ, 31.</p><p><strong>Step 2:</strong> Find the common difference. Using aₙ₊₂ = a₁ + (n+1)d: 31 = 1 + (n+1)d, so d = 30/(n+1).</p><p><strong>Step 3:</strong> Identify the 7th and (n-1)th means. The 7th mean is A₇ (the 8th term): A₈ = 1 + 7d = 1 + 210/(n+1). The (n-1)th mean is Aₙ₋₁ (the nth term): Aₙ = 1 + (n-1)d = 1 + 30(n-1)/(n+1).</p><p><strong>Step 4:</strong> Apply the ratio condition. A₈ : Aₙ = 5 : 9</p><p>[1 + 210/(n+1)] : [1 + 30(n-1)/(n+1)] = 5 : 9</p><p><strong>Step 5:</strong> Simplify. (n+1+210)/(n+1) : (n+1+30n-30)/(n+1) = 5 : 9</p><p>(n+211) : (31n-29) = 5 : 9</p><p>9(n+211) = 5(31n-29)</p><p>9n + 1899 = 155n - 145</p><p>2044 = 146n</p><p>∴ n = 14</p>
Correct Answer: 14