Find the sum of the first 40 positive integers divisible by 6

Mathematics
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find the sum of the first 40 positive integers divisible by 6

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Find the sum of the first 40 positive integers divisible by 6

Answer:
To find the sum of the first 40 positive integers divisible by 6, we need to consider the sequence of these numbers and apply the formula for the sum of an arithmetic series.

  1. Identify the Sequence:

    • The sequence of the first 40 positive integers divisible by 6 can be written as:
      6, 12, 18, 24, \ldots
    • This sequence is an arithmetic sequence where the first term (a) is 6 and the common difference (d) is also 6.

  2. Sum of an Arithmetic Series:

    • The formula for the sum of the first (n) terms of an arithmetic series is:
      S_n = \frac{n}{2} (2a + (n-1)d)

  3. Substitute the Values:

    • Here, (n = 40), (a = 6), and (d = 6). Substituting these values into the formula, we get:
      S_{40} = \frac{40}{2} \left(2 \cdot 6 + (40-1) \cdot 6\right)

  4. Simplify the Expression:

    • Calculate the terms inside the parentheses first:
      2 \cdot 6 = 12
      (40-1) \cdot 6 = 39 \cdot 6 = 234
    • Now, add these results:
      12 + 234 = 246
    • Multiply by the fraction (\frac{40}{2}):
      S_{40} = 20 \times 246 = 4920

Final Answer:
The sum of the first 40 positive integers divisible by 6 is (4920).