Sequences & Series
AP and GP
Grade 11
Question:
<p>For Problems 28–30: The numbers \(a\), \(b\), and \(c\) are between 2 and 18, such that (i) their sum is 25, (ii) the numbers 2, \(a\), and \(b\) are consecutive terms of an A.P., (iii) the numbers \(b\), \(c\), 18 are consecutive terms of a G.P.<br>The value of \(abc\) is</p>
<p>(1) 500</p>
<p>(2) 450</p>
<p>(3) 720</p>
<p>(4) 480</p>
Step-by-Step Solution
Key Concept: Use the A.P. condition (2, a, b) to express a and b in terms of common difference d, then apply the G.P. condition (b, c, 18) using c² = b·18, combined with the sum constraint a + b + c = 25 to solve the system.
<p><strong>Step 1:</strong> From A.P. condition (2, a, b): The common difference is constant.</p><p>a - 2 = b - a ⟹ a = (2 + b)/2, or equivalently: 2a = 2 + b ... (i)</p><p><strong>Step 2:</strong> From G.P. condition (b, c, 18): The common ratio is constant.</p><p>c/b = 18/c ⟹ c² = 18b ... (ii)</p><p><strong>Step 3:</strong> From sum condition: a + b + c = 25</p><p>Substituting a = (2 + b)/2 from (i):</p><p>(2 + b)/2 + b + c = 25</p><p>(2 + b)/2 + b + c = 25 ⟹ 2 + b + 2b + 2c = 50 ⟹ 3b + 2c = 48 ... (iii)</p><p><strong>Step 4:</strong> From (ii): c² = 18b ⟹ b = c²/18</p><p>Substitute into (iii): 3(c²/18) + 2c = 48</p><p>c²/6 + 2c = 48 ⟹ c² + 12c = 288 ⟹ c² + 12c - 288 = 0</p><p>(c + 24)(c - 12) = 0 ⟹ c = 12 (since c must be between 2 and 18)</p><p><strong>Step 5:</strong> Find b and a:</p><p>From (ii): b = c²/18 = 144/18 = 8</p><p>From (i): a = (2 + b)/2 = (2 + 8)/2 = 5</p><p><strong>Step 6:</strong> Verify: a + b + c = 5 + 8 + 12 = 25 ✓; A.P.: 2, 5, 8 (d = 3) ✓; G.P.: 8, 12, 18 (r = 3/2) ✓</p><p>∴ abc = 5 × 8 × 12 = <strong>480</strong></p>
Correct Answer: 4