<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 must be constant, so b/(a) = (b-2)/b. This gives a quadratic relationship that must yield natural number solutions.
<p><strong>Step 1:</strong> For a G.P. with terms a, b, b-2, the common ratio condition gives: b/a = (b-2)/b</p><p><strong>Step 2:</strong> Cross-multiply: b² = a(b-2) → b² = ab - 2a → ab - b² - 2a = 0</p><p><strong>Step 3:</strong> Rearrange: a(b-2) = b² → a = b²/(b-2)</p><p><strong>Step 4:</strong> For a ∈ ℕ, we need b-2 to divide b². Write b² = (b-2+2)² = (b-2)² + 4(b-2) + 4, so (b-2) must divide 4.</p><p><strong>Step 5:</strong> Therefore b-2 ∈ {1, 2, 4} → b ∈ {3, 4, 6}</p><p><strong>Step 6:</strong> <u>When b=3:</u> a = 9/1 = 9 ✓. Common ratio r = 3/9 = 1/3. Terms: 9, 3, 1 ✓</p><p><strong>Step 7:</strong> <u>When b=4:</u> a = 16/2 = 8 ✓. Common ratio r = 4/8 = 1/2. Terms: 8, 4, 2 ✓</p><p><strong>Step 8:</strong> <u>When b=6:</u> a = 36/4 = 9 ✓. Common ratio r = 6/9 = 2/3. Terms: 9, 6, 4 ✓</p><p><strong>Step 9:</strong> Valid pairs are (a,b) = (9,3), (8,4), (9,6). All three satisfy the G.P. condition with natural numbers.</p><p>∴ Answer: All valid pairs (A, B, C correspond to the three solutions)</p>
Correct Answer: A,B,C