Sequences & Series
Geometric Progression
Grade 11
Question:
<p>Let <em>l</em>, <em>G</em><sub>1</sub>, <em>G</em><sub>2</sub>, <em>G</em><sub>3</sub>, <em>n</em> are in GP where \(m = \dfrac{l+n}{2}\); \((l, n > 1)\). Then \((G_1)^4 + 2(G_2)^4 + (G_3)^4\) equals:</p>
<p>\(4m^2nl\)</p>
<p>\(nl(n+1)^2 = 4m^2nl\)</p>
<p>\(4m^2n^2l^2\)</p>
<p>\(nl(n-1)^2\)</p>
Step-by-Step Solution
Key Concept: In a GP with terms l, G₁, G₂, G₃, n, use the property that Gᵢ = l·rⁱ where r is common ratio. The constraint m = (l+n)/2 defines the relationship between l and n through the GP structure, allowing you to express everything in terms of l and r.
<p><strong>Step 1:</strong> Set up the GP structure. Let the common ratio be r. Then:</p><ul><li>G₁ = lr</li><li>G₂ = lr²</li><li>G₃ = lr³</li><li>n = lr⁴</li></ul><p><strong>Step 2:</strong> Use the constraint m = (l+n)/2. Since there are 5 terms in GP, the middle term satisfies: (G₂)² = l·n</p><p>This gives: (lr²)² = l·lr⁴, which is l²r⁴ = l²r⁴ ✓</p><p><strong>Step 3:</strong> Apply the AM-GM relationship. Since m = (l+n)/2 and G₂ is the geometric mean:</p><p>For a GP: G₂² = l·n, so G₂ = √(ln)</p><p><strong>Step 4:</strong> Calculate (G₁)⁴ + 2(G₂)⁴ + (G₃)⁴:</p><p>= (lr)⁴ + 2(lr²)⁴ + (lr³)⁴</p><p>= l⁴r⁴ + 2l⁴r⁸ + l⁴r¹²</p><p>= l⁴r⁴(1 + 2r⁴ + r⁸)</p><p>= l⁴r⁴(1 + r⁴)²</p><p><strong>Step 5:</strong> Note that (G₂)⁴ = l⁴r⁸ and since n = lr⁴:</p><p>(G₂)⁴ = (ln)² (from the GP mean property)</p><p>The expression simplifies to: <strong>(m⁴)</strong> or equivalently <strong>m⁴</strong></p><p>∴ Answer: <strong>m⁴</strong> (Option A)</p>
Correct Answer: A