Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>In an A.P. of which \(a\) is the first term, if the sum of the first \(p\) terms is zero, then the sum of the next \(q\) terms is</p>
<p>\(-\dfrac{a(p+q)p}{q+1}\)</p>
<p>\(\dfrac{a(p+q)p}{p+1}\)</p>
<p>\(-\dfrac{a(p+q)q}{p-1}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use the sum formula for A.P. to express the condition that S_p = 0, then find the common difference in terms of a and p, and finally calculate S_(p+q) - S_p.
<p><strong>Step 1:</strong> Let the first term be <em>a</em> and common difference be <em>d</em>.</p><p><strong>Step 2:</strong> Sum of first <em>p</em> terms: S<sub>p</sub> = p/2[2a + (p-1)d] = 0</p><p>This gives: 2a + (p-1)d = 0</p><p>∴ d = -2a/(p-1)</p><p><strong>Step 3:</strong> Sum of first (p+q) terms: S<sub>p+q</sub> = (p+q)/2[2a + (p+q-1)d]</p><p><strong>Step 4:</strong> Sum of next <em>q</em> terms = S<sub>p+q</sub> - S<sub>p</sub></p><p>= (p+q)/2[2a + (p+q-1)d] - 0</p><p><strong>Step 5:</strong> Substitute d = -2a/(p-1):</p><p>= (p+q)/2[2a + (p+q-1)·(-2a/(p-1))]</p><p>= (p+q)/2·2a[1 - (p+q-1)/(p-1)]</p><p>= (p+q)a[(p-1-p-q+1)/(p-1)]</p><p>= (p+q)a[-q/(p-1)]</p><p>= <strong>-aq(p+q)/(p-1)</strong></p><p>∴ Answer: C</p>
Correct Answer: C