What will be sum of first 40 positive integers divisible by 6

Options

A. 2460
B. 4880
C. 4920
D. 4860

Answer

To find the sum of the first 40 positive integers that are divisible by 6, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the sequence**: The first positive integer divisible by 6 is 6. The subsequent integers are 12, 18, 24, and so on. This forms an arithmetic progression (AP) where:
- First term (A) = 6
- Common difference (D) = 6

2. **Determine the number of terms (N)**: We need to find the sum of the first 40 terms, so:
- N = 40

3. **Use the formula for the sum of an arithmetic progression**: The formula for the sum of the first N terms of an AP is given by:
\[ S_N = \frac{N}{2} \times (2A + (N - 1)D) \]

4. **Substitute the values into the formula**:
- \( N = 40 \)
- \( A = 6 \)
- \( D = 6 \)

Plugging these values into the formula:
\[ S_{40} = \frac{40}{2} \times (2 \times 6 + (40 - 1) \times 6) \]

5. **Calculate the components**:
- Calculate \( \frac{40}{2} = 20 \)
- Calculate \( 2 \times 6 = 12 \)
- Calculate \( (40 - 1) = 39 \)
- Calculate \( 39 \times 6 = 234 \)

6. **Combine the results**:
\[ S_{40} = 20 \times (12 + 234) \]
\[ S_{40} = 20 \times 246 \]

7. **Final calculation**:
\[ S_{40} = 4920 \]

Thus, the sum of the first 40 positive integers divisible by 6 is **4920**.