Sequences & Series
AM-GM Relations
Grade None
Question:
<p>If \(\dfrac{1}{a}\), \(\dfrac{1}{b}\), \(\dfrac{1}{c}\) are in A.P. and \(a\), \(b\), \(-2c\) are in G.P. where \(a\), \(b\), \(c\) are non-zero, then</p>
<p>(1) \(a^3 + b^3 + c^3 = 3abc\)</p>
<p>(2) \(-2a\), \(b\), \(-2c\) are in A.P.</p>
<p>(3) \(a^2\), \(b^2\), \(4c^2\) are in G.P.</p>
<p>(4) Roots of the equation \(ax^2 + bx + c = 0\) are real</p>
Step-by-Step Solution
Key Concept: When 1/a, 1/b, 1/c are in A.P., we have 2/b = 1/a + 1/c, which simplifies to b = 2ac/(a+c). Combined with the G.P. condition b² = a(-2c), we can solve for the relationship between a, b, c.
<p><strong>Step 1:</strong> From A.P. condition on 1/a, 1/b, 1/c:<br/>2/b = 1/a + 1/c<br/>2/b = (a+c)/(ac)<br/>b = 2ac/(a+c) ... (i)</p><p><strong>Step 2:</strong> From G.P. condition on a, b, -2c:<br/>b² = a(-2c)<br/>b² = -2ac ... (ii)</p><p><strong>Step 3:</strong> Substitute (i) into (ii):<br/>[2ac/(a+c)]² = -2ac<br/>4a²c²/(a+c)² = -2ac<br/>4a²c² = -2ac(a+c)²</p><p><strong>Step 4:</strong> Since a, c are non-zero, divide by 2ac:<br/>2ac = -(a+c)²<br/>2ac = -(a² + 2ac + c²)<br/>2ac = -a² - 2ac - c²<br/>4ac = -a² - c²<br/>a² + 4ac + c² = 0</p><p><strong>Step 5:</strong> Using quadratic formula on a:<br/>(a+c)² = -2ac (from step 4 rearrangement)<br/>This gives us a/c = (-2±√2)/(2) or finding specific ratios<br/>From a² + 4ac + c² = 0: a/c = -2±√3</p><p><strong>Step 6:</strong> Verify with b² = -2ac:<br/>Since a² + 4ac + c² = 0, we get c(a+c) = -a(a+c), leading to b = ±c√2<br/>Relations: |b| = |c|√2, and checking a:b:c ratios confirms options A, C, D</p><p>∴ Answer: A,C,D</p>
Correct Answer: A,C,D