Sequences & Series
AP and GP
Grade 11
Question:
<p>For \(a, b > 0\), let \(5a - b\), \(2a + b\), \(a + 2b\) be in A.P. and \((b+1)^2\), \(ab+1\), \((a-1)^2\) are in G.P., then the value of \((a^{-1} + b^{-1})\) is ___.</p>
Step-by-Step Solution
Key Concept: Use the A.P. condition to establish a linear relationship between a and b, then apply the G.P. condition (middle term squared equals product of extremes) to solve for specific values of a and b.
<p><strong>Step 1 (A.P. Condition):</strong> If 5a - b, 2a + b, a + 2b are in A.P., then the middle term equals the average of extremes:</p><p>2(2a + b) = (5a - b) + (a + 2b)</p><p>4a + 2b = 6a + b</p><p>b = 2a ... (1)</p><p><strong>Step 2 (G.P. Condition):</strong> If (b+1)², ab+1, (a-1)² are in G.P., then:</p><p>(ab + 1)² = (b + 1)² · (a - 1)²</p><p><strong>Step 3 (Substitute b = 2a):</strong></p><p>(2a² + 1)² = (2a + 1)² · (a - 1)²</p><p>4a⁴ + 4a² + 1 = (2a + 1)²(a - 1)²</p><p>4a⁴ + 4a² + 1 = (2a² - 2a + a - 1)² · (something)</p><p><strong>Step 4 (Expand RHS):</strong></p><p>(2a + 1)²(a - 1)² = [(2a + 1)(a - 1)]² = (2a² - 2a + a - 1)² = (2a² - a - 1)²</p><p>= 4a⁴ - 4a³ - 4a² + a² + 2a + 1 = 4a⁴ - 4a³ - 3a² + 2a + 1</p><p><strong>Step 5 (Equate):</strong></p><p>4a⁴ + 4a² + 1 = 4a⁴ - 4a³ - 3a² + 2a + 1</p><p>4a³ + 7a² - 2a = 0</p><p>a(4a² + 7a - 2) = 0</p><p>Since a > 0: 4a² + 7a - 2 = 0</p><p>a = (-7 + √(49 + 32))/8 = (-7 + 9)/8 = 1/4 (taking positive root)</p><p><strong>Step 6 (Find b and final answer):</strong></p><p>b = 2a = 1/2</p><p>a⁻¹ + b⁻¹ = 4 + 2 = <strong>6</strong></p><p><em>Note: Verify a, b > 0 ✓</em></p>
Correct Answer: 1