Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant.
  • $k^{th}$ term: $a_k = a + (k-1)d$, where $a$ is the first term and $d$ is the common difference.
  • Sum of first $k$ terms: $S_k = \frac{k}{2}[2a + (k-1)d]$.

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Step-by-Step Solution

Step 1: Calculate the total savings in the first three months. The man saves ₹ 200 in each of the first three months. Saving in Month 1 = ₹ 200 Saving in Month 2 = ₹ 200 Saving in Month 3 = ₹ 200 Total saving in the first 3 months = $3 \times 200 = \text{₹ } 600$.

Step 2: Identify the savings pattern from the fourth month onwards. From the fourth month, his savings increase by ₹ 40 more than the previous month. This forms an Arithmetic Progression (AP). Saving in Month 4 = Saving in Month 3 + ₹ 40 = ₹ 200 + ₹ 40 = ₹ 240. Saving in Month 5 = Saving in Month 4 + ₹ 40 = ₹ 240 + ₹ 40 = ₹ 280. The sequence of savings from the 4th month onwards is an AP with:

• First term ($a$) = ₹ 240 (saving in the 4th month).
• Common difference ($d$) = ₹ 40.

Step 3: Set up the equation for total savings. Let $n$ be the total number of months after which the total saving reaches ₹ 11040. The total saving is the sum of savings in the first 3 months and the sum of savings in the subsequent $n-3$ months. The number of terms in the AP (from the 4th month to the $n^{th}$ month) is $k = n-3$. The sum of savings in the first 3 months is ₹ 600. The sum of savings in the AP part is $S_{n-3}$. Total Saving = (Saving in first 3 months) + (Sum of AP savings) $11040 = 600 + S_{n-3}$

Step 4: Calculate the sum of the AP savings ($S_{n-3}$). Using the formula for the sum of an AP, $S_k = \frac{k}{2}[2a + (k-1)d]$: Substitute $k = n-3$, $a = 240$, and $d = 40$: $S_{n-3} = \frac{n-3}{2}[2(240) + ((n-3)-1)40]$ $S_{n-3} = \frac{n-3}{2}[480 + (n-4)40]$

Step 5: Substitute $S_{n-3}$ into the total saving equation and solve for $n$. From Step 3: $11040 = 600 + S_{n-3}$ $11040 - 600 = S_{n-3}$ $10440 = \frac{n-3}{2}[480 + (n-4)40]$ Multiply both sides by 2: $20880 = (n-3)[480 + 40n - 160]$ $20880 = (n-3)[40n + 320]$ Factor out 40 from the bracket: $20880 = (n-3)[40(n + 8)]$ $20880 = 40(n-3)(n+8)$ Divide both sides by 40: $\frac{20880}{40} = (n-3)(n+8)$ $522 = (n-3)(n+8)$ Expand the right side: $522 = n^2 + 8n - 3n - 24$ $522 = n^2 + 5n - 24$ Rearrange into a quadratic equation: $n^2 + 5n - 24 - 522 = 0$ $n^2 + 5n - 546 = 0$

Step 6: Solve the quadratic equation. We need to find two numbers that multiply to -546 and add to 5. By factoring 546 ($546 = 2 \times 3 \times 7 \times 13$), we find the pair 26 and 21, where $26 \times (-21) = -546$ and $26 + (-21) = 5$. So, the equation can be factored as: $(n + 26)(n - 21) = 0$ This gives two possible values for $n$: $n + 26 = 0 \Rightarrow n = -26$ $n - 21 = 0 \Rightarrow n = 21$ Since the number of months cannot be negative, we discard $n = -26$. Therefore, $n = 21$ months.

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Common Mistakes & Tips

• Incorrectly defining the AP: Ensure the first term ($a$) and the number of terms ($k$) of the AP are correctly identified relative to the initial fixed savings.
• Algebraic errors: Be meticulous when expanding, simplifying, and solving the quadratic equation. Double-check calculations.
• Interpreting the solution: Always verify that the obtained value of $n$ is physically meaningful in the context of the problem (e.g., positive number of months).

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Summary

The problem was approached by separating the initial fixed savings from the subsequent savings, which followed an arithmetic progression. We calculated the sum of the fixed portion and then applied the AP sum formula to the variable portion. By setting the total savings equal to the sum of these two parts, we derived a quadratic equation. Solving this equation and selecting the positive root yielded the total number of months required to reach the target savings.

The final answer is $\boxed{21 \text{ months}}$ which corresponds to option (C).