<p>A man saves ₹200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹40 more than the saving of immediately previous month. His total saving from the start of service will be ₹11040 after</p>
Step-by-Step Solution
Key Concept: Recognize the pattern of savings and split into two parts: constant savings for first 3 months and arithmetic progression thereafter. Apply AP sum formula.
<p><strong>Step 1:</strong> The savings form an arithmetic progression pattern after the first three months.</p><p>First 3 months: ₹200, ₹200, ₹200</p><p>From month 4 onwards: ₹240, ₹280, ₹320, ... (with common difference ₹40)</p><p><strong>Step 2:</strong> Total savings = 3(200) + sum of AP starting from 240</p><p>₹600 + [sum of AP with first term 240, common difference 40, and (n-3) terms] = ₹11040</p><p><strong>Step 3:</strong> Sum of remaining AP = 11040 - 600 = 10440</p><p>Using \(S = \frac{(n-3)}{2}[2(240) + (n-4)(40)]\)</p><p>10440 = \(\frac{(n-3)}{2}[480 + 40n - 160]\)</p><p>10440 = \(\frac{(n-3)}{2}[320 + 40n]\)</p><p>20880 = (n-3)(320 + 40n)</p><p>20880 = 320n + 40n² - 960 - 120n</p><p>40n² + 200n - 21840 = 0</p><p>n² + 5n - 546 = 0</p><p>(n - 21)(n + 26) = 0</p><p>∴ n = 21 months</p>
Correct Answer: C