<p>Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178. If the first term of this AP is 10, then the median of the AP is</p>
Step-by-Step Solution
Key Concept: Use the given conditions on sum of first three terms and sum of last four terms to find the common difference and number of terms, then locate the median as the middle term(s) of the sequence.
<p><strong>Step 1:</strong> Let the AP have first term a = 10, common difference d, and n terms.</p><p><strong>Step 2:</strong> Sum of first three terms: a + (a+d) + (a+2d) = 39<br/>3a + 3d = 39<br/>3(10) + 3d = 39<br/>30 + 3d = 39<br/>d = 3</p><p><strong>Step 3:</strong> Sum of last four terms: (a + (n-4)d) + (a + (n-3)d) + (a + (n-2)d) + (a + (n-1)d) = 178<br/>4a + (4n - 10)d = 178<br/>4(10) + (4n - 10)(3) = 178<br/>40 + 12n - 30 = 178<br/>12n + 10 = 178<br/>n = 14</p><p><strong>Step 4:</strong> The AP has 14 terms (even number), so the median is the average of the 7th and 8th terms.<br/>7th term: a + 6d = 10 + 6(3) = 28<br/>8th term: a + 7d = 10 + 7(3) = 31<br/>Median = (28 + 31)/2 = 59/2 = 29.5</p><p>∴ Answer: C (29.5)</p>
Correct Answer: C