A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11,040 after:

Question

Accepted Answer

To determine why Option 3 is the correct answer, we need to calculate the man's total savings over time and see when it reaches Rs. 11,040.

1. **Initial Savings Pattern:**
   - For the first three months, the man saves Rs. 200 each month.
   - Total savings for the first three months = 3 * 200 = Rs. 600.

2. **Subsequent Savings Pattern:**
   - From the fourth month onward, his savings increase by Rs. 40 more than the previous month.
   - Therefore, his savings for the fourth month = Rs. 200 + Rs. 40 = Rs. 240.
   - For the fifth month, his savings = Rs. 240 + Rs. 40 = Rs. 280.
   - This pattern continues, forming an arithmetic sequence where the first term (a) is Rs. 240 and the common difference (d) is Rs. 40.

3. **Total Savings Calculation:**
   - We need to find the total savings after a certain number of months such that the sum equals Rs. 11,040.
   - The savings from the fourth month onward form an arithmetic series. The sum of an arithmetic series can be calculated using the formula:
     $$
     S_n = \frac{n}{2} \times (2a + (n-1)d)
     $$
   - Here, $a = 240$, $d = 40$, and $n$ is the number of months from the fourth month onward.

4. **Finding the Total Number of Months:**
   - Let $n$ be the number of months from the fourth month onward. Then the total number of months is $n + 3$.
   - The total savings from the fourth month onward is:
     $$
     S_n = \frac{n}{2} \times (2 \times 240 + (n-1) \times 40)
     $$
   - The total savings from the start is:
     $$
     600 + S_n = 11,040
     $$
   - Solving for $n$:
     $$
     600 + \frac{n}{2} \times (480 + 40n - 40) = 11,040
     $$
     $$
     600 + \frac{n}{2} \times (440 + 40n) = 11,040
     $$
     $$
     \frac{n}{2} \times (440 + 40n) = 10,440
     $$
     $$
     n \times (440 + 40n) = 20,880
     $$
     $$
     40n^2 + 440n - 20,880 = 0
     $$
   - Simplifying:
     $$
     n^2 + 11n - 522 = 0
     $$
   - Solving this quadratic equation using the quadratic formula:
     $$
     n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
     $$
     where $a = 1$, $b = 11$, $c = -522$.
   - Calculating:
     $$
     n = \frac{-11 \pm \sqrt{11^2 - 4 \times 1 \times (-522)}}{2 \times 1}
     $$
     $$
     n = \frac{-11 \pm \sqrt{121 + 2088}}{2}
     $$
     $$
     n = \frac{-11 \pm \sqrt{2209}}{2}
     $$
     $$
     n = \frac{-11 \pm 47}{2}
     $$
   - The positive solution is:
     $$
     n = \frac{36}{2} = 18
     $$
   - Therefore, the total number of months is $n + 3 = 18 + 3 = 21$.

5. **Conclusion:**
   - The man's total savings reach Rs. 11,040 after 21 months, confirming that Option 3 is correct.

**Explanation for Other Options:**
- **Option 1 (19 months):** The total savings would be less than Rs. 11,040 because the arithmetic sequence would not have accumulated enough.
- **Option 2 (20 months):** Similarly, the total savings would still be less than Rs. 11,040.
- **Option 4 (18 months):** The savings would be significantly less than Rs. 11,040, as shown in the calculations.

Thus, Option 3 is the only correct answer, as it accurately reflects the time required for the man's savings to reach Rs. 11,040.