<p>Let \(a, b, c\) (in order) be the first three terms of a sequence and satisfying \(\log\!\left(\dfrac{8b^3 - a^3 - c^3}{6abc}\right) = 0\) and \(\log b = \log(a^2 - 4) = \log(c-2)\). If \(T_n\) and \(S_n\) denote \(n^{\text{th}}\) term and sum of first \(n\) terms of the sequence, then which of the following is/are correct?</p>
Step-by-Step Solution
Key Concept: The equation 8b³ - a³ - c³ = 6abc factors as (2b - a - c)(4b² + a² + c² + 2ab + 2bc - ac) = 0, and combined with the logarithmic constraint that b = a² - 4 = c - 2, this uniquely determines an arithmetic progression with common difference 2.
<p><strong>Step 1:</strong> From log equation: log(8b³ - a³ - c³) - log(6abc) = 0, so 8b³ - a³ - c³ = 6abc.</p><p><strong>Step 2:</strong> The identity 8b³ - a³ - c³ - 6abc = (2b - a - c)(4b² + a² + c² + 2ab + 2bc - ac) implies either 2b = a + c (AP condition) or the second factor = 0 (which has no real solutions for distinct a, b, c).</p><p><strong>Step 3:</strong> From log(b) = log(a² - 4) = log(c - 2): we get b = a² - 4 and b = c - 2, so a² - 4 = c - 2, giving c = a² - 2.</p><p><strong>Step 4:</strong> Substitute into AP condition 2b = a + c: 2(a² - 4) = a + (a² - 2), which gives 2a² - 8 = a + a² - 2, so a² - a - 6 = 0, thus (a - 3)(a + 2) = 0.</p><p><strong>Step 5:</strong> For b > 0 (logarithm domain), we need a = 3. Then b = 9 - 4 = 5 and c = 9 - 2 = 7.</p><p><strong>Step 6:</strong> The sequence is 3, 5, 7, ... with first term a = 3 and common difference d = 2.</p><p><strong>Step 7:</strong> T_n = 3 + (n-1)·2 = 2n + 1 and S_n = n/2[2(3) + (n-1)·2] = n(n + 2).</p><p>∴ Answer: <strong>AP with T_n = 2n + 1 and S_n = n(n + 2)</strong></p>
Correct Answer: ABCD