<p>If \(a_1, a_2, a_3, \ldots\) are in A.P., then \(a_p, a_q, a_r\) are in A.P. if \(p\), \(q\), \(r\) are in</p>
Step-by-Step Solution
Key Concept: If a sequence is in A.P., any subsequence formed by terms at positions in A.P. will also be in A.P. This follows because the general term is linear: aₙ = a₁ + (n-1)d, and linear functions preserve arithmetic progressions.
<p><strong>Step 1:</strong> Since a₁, a₂, a₃, ... are in A.P., we can write aₙ = a₁ + (n-1)d for some common difference d.</p><p><strong>Step 2:</strong> For aₚ, aᵩ, aᵣ to be in A.P., we need: 2aᵩ = aₚ + aᵣ</p><p><strong>Step 3:</strong> Substitute the general term:</p><p>2[a₁ + (q-1)d] = [a₁ + (p-1)d] + [a₁ + (r-1)d]</p><p>2a₁ + 2(q-1)d = 2a₁ + (p-1)d + (r-1)d</p><p>2(q-1)d = (p-1)d + (r-1)d</p><p>2(q-1) = (p-1) + (r-1)</p><p>2q - 2 = p + r - 2</p><p>2q = p + r</p><p><strong>Step 4:</strong> This is the condition for p, q, r to be in A.P.</p><p>∴ Answer: A (p, q, r are in A.P.)</p>
Correct Answer: A