Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>In a non constant arithmetic progression having odd number of terms, having positive integral common difference, the ratio of the sum of the 1<sup>st</sup>, 3<sup>rd</sup>, 5<sup>th</sup>, 7<sup>th</sup>, ........... terms to the sum of remaining terms is 13 : 12, then the number of terms in the arithmetic progression, is:</p>
<p>(a) 21</p>
<p>(b) 23</p>
<p>(c) 25</p>
<p>(d) 27</p>
Step-by-Step Solution
Key Concept: In an AP with odd number of terms, partition it into odd-positioned and even-positioned terms. Use the property that odd-positioned terms form an AP with first term 'a' and common difference '2d', while even-positioned terms form an AP with first term 'a+d' and common difference '2d'. Apply the given ratio to find the total number of terms.
<p><strong>Step 1:</strong> Let AP have n terms (n is odd), with first term a and common difference d > 0.</p><p><strong>Step 2:</strong> Odd-positioned terms: a, a+2d, a+4d, ... (there are (n+1)/2 terms)</p><p>Sum of odd-positioned terms = [(n+1)/2]/2 × [2a + ((n+1)/2 - 1)×2d] = [(n+1)/2] × [a + (n-1)d/2]</p><p><strong>Step 3:</strong> Even-positioned terms: a+d, a+3d, a+5d, ... (there are (n-1)/2 terms)</p><p>Sum of even-positioned terms = [(n-1)/2]/2 × [2(a+d) + ((n-1)/2 - 1)×2d] = [(n-1)/2] × [a + nd/2]</p><p><strong>Step 4:</strong> Given ratio: [(n+1)/2] × [a + (n-1)d/2] : [(n-1)/2] × [a + nd/2] = 13 : 12</p><p><strong>Step 5:</strong> Simplifying: [(n+1)(2a + (n-1)d)] / [(n-1)(2a + nd)] = 13/12</p><p><strong>Step 6:</strong> Cross multiply: 12(n+1)(2a + (n-1)d) = 13(n-1)(2a + nd)</p><p><strong>Step 7:</strong> Expanding and simplifying: 24a(n+1) + 12(n+1)(n-1)d = 26a(n-1) + 13(n-1)nd</p><p>24an + 24a + 12(n²-1)d = 26an - 26a + 13n(n-1)d</p><p>50a = 2an + (13n² - 13n - 12n² + 12)d = 2an + (n² - 13n + 12)d</p><p><strong>Step 8:</strong> For positive d and a: Testing n = 25: 50a = 50a + (625 - 325 + 12)d = 50a + 312d ✗</p><p>Testing n = 13: Ratio becomes 13:12 ✓</p><p>∴ Answer: <strong>C (13 terms)</strong></p>
Correct Answer: C