<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: The savings follow an arithmetic progression starting from month 4 onwards. Set up two phases: fixed ₹200 for first 3 months, then AP with first term 240 and common difference 40. Use the sum formula to find the total time period.
<p><strong>Step 1:</strong> Identify the savings pattern.</p><p>Months 1-3: ₹200 each month</p><p>Month 4 onwards: ₹240, ₹280, ₹320, ... (increases by ₹40)</p><p><strong>Step 2:</strong> Calculate savings for first 3 months.</p><p>Total for months 1-3 = 3 × 200 = ₹600</p><p><strong>Step 3:</strong> Find remaining amount needed.</p><p>Remaining = 11040 - 600 = ₹10440</p><p><strong>Step 4:</strong> Apply AP sum formula for months 4 onwards.</p><p>Let n = number of months from month 4</p><p>S_n = n/2[2a + (n-1)d] where a = 240, d = 40</p><p>10440 = n/2[2(240) + (n-1)40]</p><p>10440 = n/2[480 + 40n - 40]</p><p>10440 = n/2[440 + 40n]</p><p>20880 = n[440 + 40n]</p><p>40n² + 440n - 20880 = 0</p><p>n² + 11n - 522 = 0</p><p><strong>Step 5:</strong> Solve the quadratic equation.</p><p>(n + 27)(n - 18) = 0</p><p>n = 18 (taking positive value)</p><p><strong>Step 6:</strong> Find total months.</p><p>Total months = 3 + 18 = 21 months</p><p>∴ Answer: 21 months (Option C)</p>
Correct Answer: C