Sequences & Series
AM, GM, HM
Grade 11
Question:
<p>If \(a, b, c\) are three distinct numbers in G.P., \(b, c, a\) are in A.P. and \(a, bc, abc\) are in H.P., then the possible value of \(b\) is</p>
<p>\(3+4\sqrt{2}\)</p>
<p>\(3-4\sqrt{2}\)</p>
<p>\(3+3\sqrt{2}\)</p>
<p>\(4-3\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Use the three progression conditions simultaneously: G.P. gives b²=ac, A.P. gives 2c=b+a, and H.P. gives 1/b + 1/(abc) = 2/(bc). These three equations severely constrain the system to find b in terms of a common ratio or specific value.
<p><strong>Step 1: Set up the three progression conditions</strong></p><p>Given: a, b, c are in G.P. → b² = ac ... (1)</p><p>Given: b, c, a are in A.P. → 2c = b + a ... (2)</p><p>Given: a, bc, abc are in H.P. → reciprocals in A.P.: 2/(bc) = 1/a + 1/(abc) ... (3)</p><p><strong>Step 2: Simplify the H.P. condition</strong></p><p>From (3): 2/(bc) = 1/a + 1/(abc) = (c+1)/(ac)</p><p>Cross-multiply: 2ac = bc(c+1)</p><p>Since b² = ac, substitute: 2b² = bc(c+1) → 2b = c(c+1) ... (4)</p><p><strong>Step 3: Use A.P. condition to express c in terms of a and b</strong></p><p>From (2): c = (a+b)/2 ... (5)</p><p><strong>Step 4: Substitute (5) into G.P. condition</strong></p><p>From (1): b² = a·(a+b)/2 → 2b² = a² + ab → 2b² - ab - a² = 0</p><p>Factoring: (2b + a)(b - a) = 0 → b = a or b = -a/2</p><p>Since a, b, c are distinct, b ≠ a, so b = -a/2</p><p><strong>Step 5: Verify with equation (4)</strong></p><p>From (5): c = (a - a/2)/2 = a/4</p><p>Check (4): 2(-a/2) = (a/4)(a/4 + 1) → -a = a²/16 + a/4 → -16a = a² + 4a → a² + 20a = 0 → a = -20 (a ≠ 0)</p><p>Thus b = -(-20)/2 = 10</p><p>∴ Answer: <strong>b = 10</strong> (or in some formats: b/a = -1/2 showing the ratio)</p>
Correct Answer: A