<p>The number of terms in an AP is even; the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by \(10\dfrac{1}{2}\), then the number of terms in the AP is</p>
Step-by-Step Solution
Key Concept: In an AP with even number of terms (2n), separate the sum of odd-positioned terms and even-positioned terms using the common difference to relate them. The difference between these sums gives a direct relationship with d and n.
<p><strong>Step 1:</strong> Let the AP have 2n terms: a, a+d, a+2d, ..., a+(2n-1)d</p><p><strong>Step 2:</strong> Sum of odd-positioned terms (1st, 3rd, 5th, ..., (2n-1)th):</p><p>S_odd = a + (a+2d) + (a+4d) + ... + (a+(2n-2)d) = na + d[0+2+4+...+(2n-2)]</p><p>= na + d·n(n-1) = n[a + d(n-1)] = 24</p><p><strong>Step 3:</strong> Sum of even-positioned terms (2nd, 4th, 6th, ..., 2nth):</p><p>S_even = (a+d) + (a+3d) + (a+5d) + ... + (a+(2n-1)d) = na + d[1+3+5+...+(2n-1)]</p><p>= na + d·n² = n[a + nd] = 30</p><p><strong>Step 4:</strong> From Step 2: a + d(n-1) = 24/n</p><p>From Step 3: a + nd = 30/n</p><p><strong>Step 5:</strong> Subtracting: nd - d(n-1) = (30-24)/n</p><p>d = 6/n</p><p><strong>Step 6:</strong> Last term exceeds first by 10½:</p><p>a + (2n-1)d - a = 21/2</p><p>(2n-1)d = 21/2</p><p>(2n-1)·(6/n) = 21/2</p><p>12(2n-1)/n = 21</p><p>24n - 12 = 21n</p><p>3n = 12</p><p>n = 4</p><p><strong>Step 7:</strong> Number of terms = 2n = 8</p><p>∴ Answer: B</p>
Correct Answer: B