Question

Find the sum of the first 40 positive integers divisible by 6.
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Answer 1

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Answer 2

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Ap...6, 12, 18,....

a = 6

d = a2 - a1

d = 6

nth = a + (n - 1) × d

nth = 6 + (40 - 1) × 6

nth = 6 + 39 × 6

nth = 240...

therefore, the last term of Ap is 240...

$sn \: = \frac{n}{2} (2a + (n - 1) \times d)$

$sn = \frac{40}{2} (2 \times 6 + (40 - 1) \times 6)$

$sn = 20(12 + 39 \times 6)$

$sn = 20 \times 246$

$sn = 4920$

hence, the sum of 40 integers that are divisible by 6 is 4,920...

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