A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months In each of ther mupienent montha his saving increases by Rs, 40 more than the saving of immediately previous month. His total saving s from the start of service will be Rs. 11040 after

Options

A. 21 months
B. 18 months
C. 19 months
D. 20 months

Explanation

To solve the problem step by step, we will first summarize the savings pattern and then derive the total savings equation.

### Step 1: Understand the Savings Pattern
- In the first three months, the man saves Rs. 200 each month.
- From the fourth month onwards, his savings increase by Rs. 40 more than the previous month.

### Step 2: Calculate Savings for the First Three Months
- Savings for the first three months:
- Month 1: Rs. 200
- Month 2: Rs. 200
- Month 3: Rs. 200
- Total savings for the first three months = 200 + 200 + 200 = Rs. 600.

### Step 3: Define Savings from the Fourth Month Onwards
- From the fourth month:
- Month 4: Rs. 200 + Rs. 40 = Rs. 240
- Month 5: Rs. 240 + Rs. 40 = Rs. 280
- Month 6: Rs. 280 + Rs. 40 = Rs. 320
- And so on...

### Step 4: Generalize the Savings Formula
- The savings from the fourth month can be expressed as an arithmetic progression (AP):
- First term (a) = Rs. 240 (savings in the 4th month)
- Common difference (d) = Rs. 40
- The savings for the nth month (where n > 3) can be expressed as:
\[ S_n = 240 + (n - 4) \cdot 40 \]

### Step 5: Total Savings Calculation
- The total savings after n months can be expressed as:
\[ \text{Total Savings} = 600 + \text{Sum of savings from month 4 to month n} \]
- The number of terms from month 4 to month n is (n - 3).
- The sum of an AP is given by:
\[ S = \frac{n}{2} \cdot (2a + (n - 1)d) \]
- Here, \( n = n - 3 \), \( a = 240 \), and \( d = 40 \):
\[ S = \frac{(n - 3)}{2} \cdot (2 \cdot 240 + (n - 4) \cdot 40) \]

### Step 6: Set Up the Equation
- The total savings equation becomes:
\[ 600 + \frac{(n - 3)}{2} \cdot (480 + (n - 4) \cdot 40) = 11040 \]
- Simplifying gives:
\[ \frac{(n - 3)}{2} \cdot (480 + 40n - 160) = 11040 - 600 \]
\[ \frac{(n - 3)}{2} \cdot (40n + 320) = 10440 \]

### Step 7: Solve for n
- Multiply both sides by 2:
\[ (n - 3)(40n + 320) = 20880 \]
- Expanding and rearranging gives:
\[ 40n^2 + 320n - 120n - 960 = 20880 \]
\[ 40n^2 + 200n - 21840 = 0 \]
- Dividing through by 40:
\[ n^2 + 5n - 546 = 0 \]

### Step 8: Factor or Use Quadratic Formula
- Factoring gives:
\[ (n - 21)(n + 26) = 0 \]
- Thus, \( n = 21 \) or \( n = -26 \). Since n cannot be negative, we have:
\[ n = 21 \]

### Final Answer
The man saves for a total of **21 months**.