Find the sum of the first 40 positive integers divisible by 6.
\(4920\)
Step 1: Identify the AP.
The positive integers divisible by 6 form the sequence:
\[6, 12, 18, 24, \ldots\]
This is an AP with:
First term: \(a = 6\)
Common difference: \(d = 6\)
Step 2: We need the sum of the first 40 terms.
Use the sum formula for an AP:
\[S_n = \frac{n}{2}[2a + (n - 1)d]\]
Substitute \(n = 40, a = 6, d = 6\):
\[S_{40} = \frac{40}{2}[2(6) + 39(6)]\]
Simplify inside the bracket:
\[2(6) = 12\]
\[39(6) = 234\]
So,
\[S_{40} = 20(12 + 234) = 20 \times 246\]
Step 3: Multiply to get the final answer.
\[20 \times 246 = 4920\]
Conclusion: The sum of the first 40 positive integers divisible by 6 is 4920.