The maximum sum of the A.P. series 40,36,32 . . .is

Options

A. can't be determined
B. 20
C. 220
D. 232

Answer

To find the maximum sum of the arithmetic progression (A.P.) series 40, 36, 32, ..., we can follow these steps:

### Step-by-Step Solution:

1. **Identify the A.P. Series**:
The given series is 40, 36, 32, ... This is an arithmetic progression where the first term \( a = 40 \) and the common difference \( d = 36 - 40 = -4 \).

2. **Determine the Terms**:
The series continues with negative terms. The next terms will be:
- 28 (40 - 12)
- 24 (40 - 16)
- 20 (40 - 20)
- 16 (40 - 24)
- 12 (40 - 28)
- 8 (40 - 32)
- 4 (40 - 36)
- 0 (40 - 40)
- -4 (40 - 44)
- ...

We can see that the terms will eventually become negative.

3. **Finding the Maximum Sum**:
To find the maximum sum, we need to stop before we reach the negative terms. The last positive term is 4. Therefore, we will sum the series from 40 down to 4.

4. **Sum of the Series**:
The series we will sum is:
\[ S = 40 + 36 + 32 + 28 + 24 + 20 + 16 + 12 + 8 + 4 \]

5. **Factor Out Common Terms**:
We can factor out 4 from each term:
\[ S = 4 \times (10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) \]

6. **Use the Formula for Sum of First n Natural Numbers**:
The sum of the first \( n \) natural numbers is given by the formula:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
Here, \( n = 10 \):
\[ \text{Sum} = \frac{10 \times (10 + 1)}{2} = \frac{10 \times 11}{2} = 55 \]

7. **Calculate the Final Sum**:
Now, substituting back into our equation for \( S \):
\[ S = 4 \times 55 = 220 \]

8. **Conclusion**:
The maximum sum of the A.P. series 40, 36, 32, ... is **220**.