<p>Let \(a\), \(b\) and \(b-2\) are the first three terms (in order) of a G.P. where \(a, b \in N\). Identify which of the following statement(s) is(are) correct?</p>
<p>If \(a \in [1, 8]\), then \(r = \dfrac{1}{2}\)</p>
<p>If \(a \in [1, 8]\), then \(S_\infty = 16\)</p>
<p>If \(a \in (8, 11]\), then \(S_\infty\) can be equal to 27</p>
<p>If \(a \in (1, 8]\), then \(S_\infty = \dfrac{27}{2}\)</p>
Step-by-Step Solution
Key Concept: In a G.P., the ratio between consecutive terms is constant, so b/a = (b-2)/b. This gives a quadratic in b that determines the valid natural number solutions for (a,b).
<p><strong>Step 1:</strong> Use the G.P. condition. For a, b, (b-2) in G.P., the common ratio r must satisfy:</p><p>b/a = (b-2)/b</p><p><strong>Step 2:</strong> Cross-multiply: b² = a(b-2)</p><p>Rearranging: b² = ab - 2a, so b² - ab + 2a = 0</p><p><strong>Step 3:</strong> Solve for a: 2a = b² - ab, giving a(b-2) = b²</p><p>Therefore: a = b²/(b-2)</p><p><strong>Step 4:</strong> For a ∈ ℕ, (b-2) must divide b². Write b = (b-2) + 2:</p><p>b² = [(b-2) + 2]² = (b-2)² + 4(b-2) + 4</p><p>So: a = [(b-2)² + 4(b-2) + 4]/(b-2) = (b-2) + 4 + 4/(b-2)</p><p><strong>Step 5:</strong> For a ∈ ℕ, (b-2) must divide 4. Thus b-2 ∈ {1, 2, 4}</p><p>This gives b ∈ {3, 4, 6}</p><p><strong>Step 6:</strong> Check each case:</p><p>• b = 3: a = 1 + 4 + 4 = 9. G.P. is 9, 3, 1 with r = 1/3 ✓</p><p>• b = 4: a = 2 + 4 + 2 = 8. G.P. is 8, 4, 2 with r = 1/2 ✓</p><p>• b = 6: a = 4 + 4 + 1 = 9. G.P. is 9, 6, 4 with r = 2/3 ✓</p><p>∴ The valid pairs are (a,b) = (9,3), (8,4), (9,6). Statements depending on these solutions are correct.</p>
Correct Answer: A,B,C