Sequences & Series
A.P. and G.P.
Grade 11
Question:
<p>If \(x^2 + 3x + 2\), \(x^2 - x - 10\) and \(x^2 + x - 4 + y^2\) are in A.P. as well as in G.P., then find the values of \(x\) and \(y\), and the common value of each term.</p>
<p>\(x = -3,\ y = 0\)</p>
<p>Each number equals 2</p>
<p>\(x = 3,\ y = 0\)</p>
<p>Each number equals \(-2\)</p>
Step-by-Step Solution
Key Concept: If three terms are in both A.P. and G.P. simultaneously, they must all be equal (since A.P. requires 2b = a+c while G.P. requires b² = ac, which together force a = b = c). Use this constraint to set up equations.
<p><strong>Step 1: Use the simultaneity condition</strong></p><p>Let a = x² + 3x + 2, b = x² - x - 10, c = x² + x - 4 + y²</p><p>If three terms are in both A.P. and G.P., then: a = b = c</p><p><strong>Step 2: Apply a = b</strong></p><p>x² + 3x + 2 = x² - x - 10</p><p>3x + 2 = -x - 10</p><p>4x = -12</p><p>x = -3</p><p><strong>Step 3: Find the common value using a = b</strong></p><p>When x = -3: a = (-3)² + 3(-3) + 2 = 9 - 9 + 2 = 2</p><p>Verify: b = (-3)² - (-3) - 10 = 9 + 3 - 10 = 2 ✓</p><p><strong>Step 4: Apply b = c to find y</strong></p><p>x² - x - 10 = x² + x - 4 + y²</p><p>-x - 10 = x - 4 + y²</p><p>With x = -3: -(-3) - 10 = -3 - 4 + y²</p><p>3 - 10 = -7 + y²</p><p>-7 = -7 + y²</p><p>y² = 0, so y = 0</p><p><strong>Step 5: Verification</strong></p><p>All three terms equal: a = b = c = 2 ✓</p><p>∴ <strong>x = -3, y = 0, common value = 2</strong></p>
Correct Answer: A and B