Sequences & Series
Arithmetic Progressions
Grade 11
Question:
<p>Let \(\{A_n\}\) and \(\{B_n\}\) be two arithmetic progressions with \(\dfrac{A_1}{B_1} = \dfrac{a_1}{b_1} = \dfrac{1-1}{2} = 0\), \(\dfrac{A_2}{B_2} = \dfrac{1}{4}\), and \(\dfrac{A_3}{B_3} = \dfrac{1}{3}\). Find the value of \(\dfrac{a_3+a_5+a_7}{3(b_3+b_9)} + \dfrac{a_4+a_{10}}{2(b_2+b_{10})}\).</p>
<p>A) \(\dfrac{3}{11}\)</p>
<p>B) \(\dfrac{4}{11}\)</p>
<p>C) \(\dfrac{7}{11}\)</p>
<p>D) \(\dfrac{5}{11}\)</p>
Step-by-Step Solution
Key Concept: For arithmetic progressions, use the property that the sum of terms equidistant from the center equals twice the middle term. Express ratios of sums using the common difference relationships derived from given conditions.
<p><strong>Step 1: Set up the APs</strong></p><p>Let {aₙ} have first term a and common difference d. Let {bₙ} have first term b and common difference e.</p><p>Then aₙ = a + (n-1)d and bₙ = b + (n-1)e</p><p><strong>Step 2: Use given conditions</strong></p><p>From a₁/b₁ = 0: a/b = 0, so a = 0</p><p>From a₂/b₂ = 1/4: d/(b+e) = 1/4, so 4d = b+e ... (i)</p><p>From a₃/b₃ = 1/3: 2d/(b+2e) = 1/3, so 6d = b+2e ... (ii)</p><p><strong>Step 3: Solve for b and e in terms of d</strong></p><p>Subtracting (i) from (ii): 2d = e, so e = 2d</p><p>Substituting into (i): 4d = b + 2d, so b = 2d</p><p><strong>Step 4: Express the sequences</strong></p><p>aₙ = (n-1)d and bₙ = 2d + (n-1)(2d) = 2nd</p><p><strong>Step 5: Calculate first fraction</strong></p><p>a₃ + a₅ + a₇ = 2d + 4d + 6d = 12d</p><p>b₃ + b₉ = 6d + 18d = 24d</p><p>First fraction = 12d / [3(24d)] = 12d / 72d = 1/6</p><p><strong>Step 6: Calculate second fraction</strong></p><p>a₄ + a₁₀ = 3d + 9d = 12d</p><p>b₂ + b₁₀ = 4d + 20d = 24d</p><p>Second fraction = 12d / [2(24d)] = 12d / 48d = 1/4</p><p><strong>Step 7: Final answer</strong></p><p>1/6 + 1/4 = 2/12 + 3/12 = 5/12</p><p>∴ Answer: D</p>
Correct Answer: D