The sum of first 40 positive integers divisible by 6 is

Options

A. 2460
B. 3640
C. 4920
D. 4860

Explanation

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

### Step 1: Identify the Sequence
The first 40 positive integers divisible by 6 form an arithmetic progression (AP):
- The first term \( a = 6 \)
- The common difference \( d = 6 \)

The sequence of the first 40 terms is:
\[ 6, 12, 18, 24, \ldots, 240 \]

### Step 2: Determine the Number of Terms
We have \( n = 40 \) (the number of terms).

### Step 3: Use the Formula for the Sum of an AP
The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

### Step 4: Substitute the Values
Substituting the values into the formula:
- \( n = 40 \)
- \( a = 6 \)
- \( d = 6 \)

We get:
\[ S_{40} = \frac{40}{2} \times (2 \times 6 + (40 - 1) \times 6) \]

### Step 5: Simplify the Expression
Calculating step by step:
1. Calculate \( \frac{40}{2} = 20 \)
2. Calculate \( 2 \times 6 = 12 \)
3. Calculate \( 40 - 1 = 39 \)
4. Calculate \( 39 \times 6 = 234 \)
5. Now substitute back:
\[ S_{40} = 20 \times (12 + 234) \]
6. Calculate \( 12 + 234 = 246 \)
7. Finally, calculate \( 20 \times 246 \)

### Step 6: Final Calculation
\[ 20 \times 246 = 4920 \]

### Conclusion
The sum of the first 40 positive integers divisible by 6 is \( 4920 \).

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