<p>Given \(a\), \(b\) and \(c\) be three distinct real numbers in G.P. and \(a + b + c = xb\). Let \(r\) be the common ratio of the G.P. If \(x\) can take values, then which of the following is correct?</p>
<p>\(x \in (-\infty, -1) \cup (3, \infty)\)</p>
<p>\(x \in (-\infty, -1) \cup (3, \infty)\)</p>
<p>\(x \in (-1, 3)\)</p>
<p>\(x \in (-3, 1)\)</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in G.P. with common ratio r, we have b = a/r and c = ar. Expressing the sum condition a + b + c = xb in terms of a single variable and r, then analyzing the constraint that x must be real for distinct values.
<p><strong>Step 1:</strong> Let the G.P. have first term a and common ratio r. Then b = ar and c = ar².</p><p><strong>Step 2:</strong> Use the condition a + b + c = xb:<br>a + ar + ar² = x(ar)<br>Dividing by a (a ≠ 0): 1 + r + r² = xr</p><p><strong>Step 3:</strong> Rearrange to form a quadratic in r:<br>r² + r(1 - x) + 1 = 0</p><p><strong>Step 4:</strong> For r to be real and distinct from 1 (distinctness requirement), the discriminant must be non-negative and r ≠ 1:<br>Δ = (1 - x)² - 4 ≥ 0<br>(1 - x)² ≥ 4<br>|1 - x| ≥ 2<br>1 - x ≥ 2 or 1 - x ≤ -2<br>x ≤ -1 or x ≥ 3</p><p><strong>Step 5:</strong> Verify r = 1 doesn't satisfy our domain: If r = 1, then 1 + 1(1-x) + 1 = 0 → x = 3. Check: when x = 3, r² - 2r + 1 = 0 → (r-1)² = 0 → r = 1 (duplicate root, violates distinctness).</p><p><strong>Step 6:</strong> Therefore x ∈ (-∞, -1] ∪ [3, ∞), excluding x = 3:<br>x ∈ (-∞, -1] ∪ (3, ∞)</p><p>∴ Answer: A</p>
Correct Answer: A