Sequences & Series
Infinite GP and AP/HP relations
Grade 11
Question:
<p>We have \(x = \displaystyle\sum_{n=0}^{\infty} a^n\), \(y = \displaystyle\sum_{n=0}^{\infty} b^n\), \(z = \displaystyle\sum_{n=0}^{\infty} c^n\), where \(a, b, c\) are in AP and \(|a| < 1\), \(|b| < 1\), \(|c| < 1\). Then \(\dfrac{1}{x}\), \(\dfrac{1}{y}\), \(\dfrac{1}{z}\) are in:</p>
<p>AP</p>
<p>GP</p>
<p>HP</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in AP, we have b = (a+c)/2. Use this constraint along with the geometric series formula to establish a relationship between x, y, z that must hold regardless of the specific values of a, b, c.
<p><strong>Step 1:</strong> Apply the geometric series formula. Since |a|, |b|, |c| < 1:</p><p>x = 1/(1-a), y = 1/(1-b), z = 1/(1-c)</p><p><strong>Step 2:</strong> Use the AP condition. Since a, b, c are in AP:</p><p>b = (a+c)/2, which means 2b = a + c</p><p><strong>Step 3:</strong> Express the relationship in terms of x, y, z. From the formulas above:</p><p>a = 1 - 1/x, b = 1 - 1/y, c = 1 - 1/z</p><p><strong>Step 4:</strong> Substitute into 2b = a + c:</p><p>2(1 - 1/y) = (1 - 1/x) + (1 - 1/z)</p><p>2 - 2/y = 2 - 1/x - 1/z</p><p>2/y = 1/x + 1/z</p><p><strong>Step 5:</strong> This gives us the key relationship:</p><p>y = 2xz/(x+z)</p><p>Or equivalently: 2/y = 1/x + 1/z, meaning y is the harmonic mean of x and z.</p><p>∴ Answer: C</p>
Correct Answer: C