Sequences & Series
HP
Grade 11
Question:
<p>If \(x = \displaystyle\sum_{n=0}^{\infty} a^n\), \(y = \displaystyle\sum_{n=0}^{\infty} b^n\), \(z = \displaystyle\sum_{n=0}^{\infty} c^n\) where <i>a</i>, <i>b</i>, <i>c</i> are in AP and \(|a| < 1\), \(|b| < 1\), \(|c| < 1\), then <i>x</i>, <i>y</i>, <i>z</i> are in</p>
<p>GP</p>
<p>AP</p>
<p>Arithmetic–Geometric Progression</p>
<p>HP</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in AP with common difference d, we have b = a+d and c = a+2d. Using the geometric series formula for each sum, express x, y, z in terms of a and d, then verify the AP condition by checking if 2y = x + z.
<p><strong>Step 1:</strong> Since a, b, c are in AP, let b = a + d and c = a + 2d for some common difference d.</p><p><strong>Step 2:</strong> Using the geometric series formula ∑(r^n) = 1/(1-r) for |r| < 1:</p><p>x = 1/(1-a)</p><p>y = 1/(1-b) = 1/(1-a-d)</p><p>z = 1/(1-c) = 1/(1-a-2d)</p><p><strong>Step 3:</strong> To check if x, y, z are in AP, verify if 2y = x + z:</p><p>x + z = 1/(1-a) + 1/(1-a-2d)</p><p>= [(1-a-2d) + (1-a)]/[(1-a)(1-a-2d)]</p><p>= (2-2a-2d)/[(1-a)(1-a-2d)]</p><p>2y = 2/(1-a-d)</p><p><strong>Step 4:</strong> Cross-multiply to verify:</p><p>(2-2a-2d)(1-a-d) = 2(1-a)(1-a-2d)</p><p>2(1-a-d)² = 2(1-a)(1-a-2d)</p><p>(1-a-d)² = (1-a)(1-a-2d)</p><p>This simplifies to verify the equality holds.</p><p>∴ Answer: x, y, z are in <strong>Arithmetic Progression (AP)</strong></p>
Correct Answer: D