Sequences & Series
AM and GM
Grade 11
Question:
<p><strong>34.</strong> If \(a, b\) and \(c\) are in A.P., and \(p\) and \(p'\) are, respectively, A.M. and G.M. between \(a\) and \(b\) while \(q, q'\) are, respectively, the A.M. and G.M. between \(b\) and \(c\), then</p>
<p>\(p^2 + q^2 = p'^2 + q'^2\)</p>
<p>\(pq = p'q'\)</p>
<p>\(p^2 - q^2 = p'^2 - q'^2\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in A.P., we have b = (a+c)/2. Use this constraint along with the definitions of A.M. and G.M. for the pairs (a,b) and (b,c) to establish relationships between p, p', q, q'.
<p><strong>Step 1:</strong> Since a, b, c are in A.P., we have: b = (a+c)/2, which gives 2b = a + c</p><p><strong>Step 2:</strong> Define the means explicitly:<br>• p = A.M. between a and b: p = (a+b)/2<br>• p' = G.M. between a and b: p' = √(ab)<br>• q = A.M. between b and c: q = (b+c)/2<br>• q' = G.M. between b and c: q' = √(bc)</p><p><strong>Step 3:</strong> Add the A.M.s:<br>p + q = (a+b)/2 + (b+c)/2 = (a+2b+c)/2</p><p><strong>Step 4:</strong> Use the A.P. condition 2b = a + c:<br>p + q = (a+c+2b)/2 = (2b+2b)/2 = 2b</p><p><strong>Step 5:</strong> Multiply the G.M.s:<br>p'q' = √(ab) · √(bc) = √(ab²c) = b√(ac)</p><p><strong>Step 6:</strong> From 2b = a + c, by A.M.-G.M. inequality: b = (a+c)/2 ≥ √(ac)<br>Therefore: p'q' = b√(ac) and p + q = 2b</p><p>∴ The relationship <strong>p + q = 2b</strong> holds, and typically <strong>p'q' ≤ b²</strong> with equality when a = c</p>
Correct Answer: C