Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(a\), \(\dfrac{1}{b}\), \(c\) and \(\dfrac{1}{p}\), \(q\), \(\dfrac{1}{r}\) form two arithmetic progressions of the same common difference, then \(a\), \(q\), \(c\) are in A.P. if</p>
<p>\(p\), \(b\), \(r\) are in A.P.</p>
<p>\(\dfrac{1}{p}\), \(\dfrac{1}{b}\), \(\dfrac{1}{r}\) are in A.P.</p>
<p>\(p\), \(b\), \(r\) are in G.P.</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: If two APs have the same common difference d, then a = 1/b - d, c = 1/b + d, and 1/p = q - d, 1/r = q + d. Use these relationships and the condition that a, q, c are in AP to find the constraint.
<p><strong>Step 1:</strong> Set up the AP conditions. For the first AP with common difference d: a, 1/b, c means 1/b - a = d and c - 1/b = d.</p><p>This gives: <strong>a = 1/b - d</strong> and <strong>c = 1/b + d</strong></p><p><strong>Step 2:</strong> For the second AP with the same common difference d: 1/p, q, 1/r means q - 1/p = d and 1/r - q = d.</p><p>This gives: <strong>q = 1/p + d</strong> and <strong>1/r = q + d</strong></p><p><strong>Step 3:</strong> For a, q, c to be in AP, we need: q - a = c - q, or equivalently 2q = a + c.</p><p><strong>Step 4:</strong> Substitute: 2q = a + c becomes 2(1/p + d) = (1/b - d) + (1/b + d)</p><p>⟹ 2/p + 2d = 2/b</p><p>⟹ <strong>2/p = 2/b</strong> (the 2d cancels)</p><p>⟹ <strong>p = b</strong></p><p><strong>Step 5:</strong> Verify: If p = b, then q = 1/b + d, a = 1/b - d, c = 1/b + d.</p><p>Check: a + c = (1/b - d) + (1/b + d) = 2/b and 2q = 2(1/b + d) = 2/b + 2d... Wait, recalculate: when p = b, we get 2/p = 2/b, so 2q = 2/b.</p><p>∴ Answer: <strong>p = b</strong></p>
Correct Answer: A