Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let \(a_1, a_2, a_3, \ldots\) be terms of an AP. If \(\dfrac{a_1 + a_2 + \cdots + a_p}{a_1 + a_2 + \cdots + a_q} = \dfrac{p^2}{q^2},\ p \neq q\), then \(\dfrac{a_6}{a_{21}}\) equals</p>
<p>\(\dfrac{41}{11}\)</p>
<p>\(\dfrac{7}{2}\)</p>
<p>\(\dfrac{2}{7}\)</p>
<p>\(\dfrac{11}{41}\)</p>
Step-by-Step Solution
Key Concept: Use the sum formula for AP: S_n = n/2[2a + (n-1)d], then express the ratio of sums in terms of a and d. The given ratio condition uniquely determines the relationship between a and d, allowing us to find individual terms.
<p><strong>Step 1:</strong> For AP with first term a and common difference d:</p><p>S_p = p/2[2a + (p-1)d] and S_q = q/2[2a + (q-1)d]</p><p><strong>Step 2:</strong> Form the ratio:</p><p>S_p/S_q = [p(2a + (p-1)d)]/[q(2a + (q-1)d)] = p²/q²</p><p><strong>Step 3:</strong> Cross multiply:</p><p>q·p(2a + (p-1)d) = p·q(2a + (q-1)d)·(p/q)²</p><p>This simplifies to: q(2a + (p-1)d) = (p/q)(2a + (q-1)d)</p><p><strong>Step 4:</strong> Expanding: 2aq + (p-1)dq = 2ap/q + (q-1)dp/q</p><p>Multiply by q: 2aq² + (p-1)dq² = 2ap + (q-1)dp</p><p><strong>Step 5:</strong> Rearranging: 2a(q² - p) = d[p(q-1) - q(p-1)]</p><p>Simplifying RHS: d[pq - p - pq + q] = d(q - p)</p><p>Therefore: 2a(q² - p) = d(q - p)</p><p><strong>Step 6:</strong> For this to hold for all valid p, q: <strong>2a = d</strong></p><p><strong>Step 7:</strong> Now find a₆ and a₂₁:</p><p>a₆ = a + 5d = a + 5(2a) = 11a</p><p>a₂₁ = a + 20d = a + 20(2a) = 41a</p><p><strong>Step 8:</strong> Therefore: a₆/a₂₁ = 11a/41a = 11/41</p><p>∴ Answer: D (11/41)</p>
Correct Answer: D