Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p><strong>21.</strong> <em>ABC</em> is a right-angled triangle in which \(\angle B = 90°\) and \(BC = a\). If \(n\) points \(L_1, L_2, \ldots, L_n\) on \(AB\) is divided in \(n+1\) equal parts and \(L_1M_1, L_2M_2, \ldots, L_nM_n\) are line segments parallel to \(BC\) and \(M_1, M_2, \ldots, M_n\) are on \(AC\), then the sum of the lengths of \(L_1M_1, L_2M_2, \ldots, L_nM_n\) is</p>
<p>\(\dfrac{a(n+1)}{2}\)</p>
<p>\(\dfrac{a(n-1)}{2}\)</p>
<p>\(\dfrac{an}{2}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Since AB is divided into n+1 equal parts by points L₁, L₂, ..., Lₙ, and segments LᵢMᵢ are parallel to BC, by similar triangles each LᵢMᵢ forms a constant ratio with BC. The sum becomes an arithmetic series with first term a/(n+1) and common difference a/(n+1).
<p><strong>Step 1: Set up coordinate system</strong></p><p>Place B at origin, BC along positive x-axis (length a), and BA along positive y-axis. Since ∠B = 90°, triangle ABC has BC ⊥ AB.</p><p><strong>Step 2: Locate points on AB</strong></p><p>Points L₁, L₂, ..., Lₙ divide AB into n+1 equal parts. If AB has length h, then:</p><p>• L₁ is at distance h/(n+1) from B</p><p>• Lᵢ is at distance i·h/(n+1) from B</p><p><strong>Step 3: Apply similar triangles</strong></p><p>Since LᵢMᵢ ∥ BC, triangles ALᵢMᵢ and ABC are similar. The ratio of similarity is:</p><p>AL₁/AB = (h - i·h/(n+1))/h = (n+1-i)/(n+1)</p><p>Therefore: LᵢMᵢ = BC × (n+1-i)/(n+1) = a(n+1-i)/(n+1)</p><p><strong>Step 4: Reindex and sum</strong></p><p>For i = 1 to n:</p><p>• L₁M₁ = an/(n+1)</p><p>• L₂M₂ = a(n-1)/(n+1)</p><p>• ...</p><p>• LₙMₙ = a/(n+1)</p><p><strong>Step 5: Calculate total sum</strong></p><p>Sum = a/(n+1) × [n + (n-1) + (n-2) + ... + 1]</p><p>Sum = a/(n+1) × n(n+1)/2</p><p>Sum = <strong>an/2</strong></p><p>∴ Answer: C</p>
Correct Answer: C