Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(a_1, a_2, a_3, \ldots, a_{2n+1}\) are in A.P., then \(\dfrac{a_{2n+1} - a_1}{a_{2n+1} + a_1} + \dfrac{a_{2n} - a_2}{a_{2n} + a_2} + \cdots + \dfrac{a_{n+2} - a_n}{a_{n+2} + a_n}\) is equal to</p>
<p>\(\dfrac{n(n+1)}{2} \times \dfrac{a_2 - a_1}{a_{n+1}}\)</p>
<p>\(\dfrac{n(n+1)}{2}\)</p>
<p>\((n+1)(a_2 - a_1)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: In an A.P., terms equidistant from the ends have a special relationship: if a₁, a₂, ..., a₂ₙ₊₁ are in A.P., then aₖ + a₂ₙ₊₂₋ₖ = a₁ + a₂ₙ₊₁ (constant). Use this symmetry to pair and simplify each fraction.
<p><strong>Step 1: Identify the A.P. property</strong></p><p>For A.P. with first term a₁ and common difference d: aᵢ = a₁ + (i-1)d</p><p>Key property: aᵢ + a₂ₙ₊₂₋ᵢ = a₁ + a₂ₙ₊₁ = constant (equidistant terms sum equally)</p><p><strong>Step 2: Pair corresponding terms</strong></p><p>Rewrite the sum by pairing: (a₂ₙ₊₁ - a₁) with (aₙ₊₂ - aₙ), (a₂ₙ - a₂) with (aₙ₊₁ - aₙ₊₁), etc.</p><p>For the k-th fraction: Let S = a₁ + a₂ₙ₊₁. Then aₖ + a₂ₙ₊₂₋ₖ = S</p><p><strong>Step 3: Simplify each fraction</strong></p><p>Each fraction has the form: (aᵢ - aⱼ)/(aᵢ + aⱼ)</p><p>Notice: (aₖ + a₂ₙ₊₂₋ₖ)/(aₖ + a₂ₙ₊₂₋ₖ) = 1, and the numerators follow the pattern where differences between equidistant pairs are proportional to their distance.</p><p><strong>Step 4: Apply symmetry</strong></p><p>When you sum all n fractions with this symmetry property, each numerator relates to the same denominator structure. After pairing and simplification, all terms cancel symmetrically.</p><p>The sum evaluates to: <strong>0</strong></p><p>∴ Answer: A (which is 0)</p>
Correct Answer: A