Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let \(a_1, a_2, a_3, \ldots\) be terms of an A.P. If \(\frac{a_1 + a_2 + \ldots + a_p}{a_1 + a_2 + \ldots + a_q} = \frac{p^2}{q^2}\), \(p \neq q\), then \(\frac{a_p}{a_q}\) equals</p>
<p>(A) \(\frac{p}{q}\)</p>
<p>(B) \(\frac{2p-1}{2q-1}\)</p>
<p>(C) \(\frac{p^2}{q^2}\)</p>
<p>(D) \(\frac{p+q}{p-q}\)</p>
Step-by-Step Solution
Key Concept: Use the sum formula for A.P. and the given ratio to establish a relationship between the first term and common difference, then find the ratio of individual terms.
<p><strong>Solution:</strong> For an A.P. with first term \(a\) and common difference \(d\):</p><p>Sum of first \(n\) terms: \(S_n = \frac{n}{2}(2a + (n-1)d)\)</p><p>Given: \(\frac{S_p}{S_q} = \frac{p^2}{q^2}\)</p><p>\(\frac{\frac{p}{2}(2a + (p-1)d)}{\frac{q}{2}(2a + (q-1)d)} = \frac{p^2}{q^2}\)</p><p>\(\frac{p(2a + (p-1)d)}{q(2a + (q-1)d)} = \frac{p^2}{q^2}\)</p><p>\(\frac{2a + (p-1)d}{2a + (q-1)d} = \frac{p}{q}\)</p><p>Cross-multiplying and solving: \(q(2a + (p-1)d) = p(2a + (q-1)d)\)</p><p>\(2aq + pqd - qd = 2ap + pqd - pd\)</p><p>\(2a(q-p) = d(q - p)\) ... if \(q \neq p\), then \(2a = d\) (after careful algebra)</p><p>Actually: \(a_p = a + (p-1)d\) and \(a_q = a + (q-1)d\)</p><p>\(\frac{a_p}{a_q} = \frac{a + (p-1)d}{a + (q-1)d} = \frac{2p-1}{2q-1}\) (with \(a = \frac{d}{2}\))</p><p>∴ Answer is B.</p>
Correct Answer: B