Find the sum of the first 40 positive integers divisible by 6.
Step-by-Step Solution
Key Concept: The numbers form an arithmetic progression (AP) with first term $a=6$ and common difference $d=6$. Use the sum formula for the first $n$ terms of an AP: $S_n = \frac{n}{2}[2a+(n-1)d]$ or $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term.
1. Identify the AP:
\[ a = 6, \quad d = 6 \]
2. Number of terms required: $n = 40$.
3. Find the last term $l$ of the 40th term:
\[ l = a + (n-1)d = 6 + (40-1)\times6 = 6 + 39\times6 = 6 + 234 = 240 \]
4. Apply the sum formula using the first and last terms:
\[ S_{40} = \frac{n}{2}(a + l) = \frac{40}{2}(6 + 240) \]
\[ S_{40} = 20 \times 246 = 4920 \]
Alternatively, using the general AP sum formula:
\[ S_{40} = \frac{40}{2}[2\times6 + (40-1)\times6] = 20[12 + 234] = 20\times246 = 4920 \]
5. Hence, the sum of the first 40 positive integers divisible by 6 is $\boxed{4920}$.
Correct Answer: 4920