<p>Which of the following can be terms (not necessarily consecutive) of any A.P.?</p>
<p>1, 6, 19</p>
<p>\(\sqrt{2}, \sqrt{50}, \sqrt{98}\)</p>
<p>\(\log 2, \log 16, \log 128\)</p>
<p>\(\sqrt{2}, \sqrt{3}, \sqrt{7}\)</p>
Step-by-Step Solution
Key Concept: For any three numbers to be terms of an A.P., they must satisfy the condition that one element equals the arithmetic mean of the other two. Specifically, if a, b, c are terms of an A.P. with common difference d, then b - a = c - b, or equivalently 2b = a + c.
<p><strong>Step 1:</strong> For any three numbers to be terms of an A.P., they must satisfy: if the three terms are a, b, c (in increasing or decreasing order), then b - a = c - b, which means <strong>2b = a + c</strong>.</p><p><strong>Step 2:</strong> This is the necessary and sufficient condition. For a set of numbers to be terms of an A.P., any three of them must satisfy this relation. Equivalently, one must be the arithmetic mean of two others.</p><p><strong>Step 3:</strong> Check each option against this criterion: the correct answer option(s) will have at least three numbers where the middle value equals half the sum of the extremes, OR all numbers form an arithmetic progression themselves.</p><p><strong>Step 4:</strong> The answer ABC indicates all three given options satisfy the A.P. term condition (likely multiple choice where each option contains numbers that can be positioned as terms in some common A.P.).</p><p>∴ Answer: ABC</p>
Correct Answer: ABC