Question

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

Answer 1

a = 6

n = 40

d = 6

$\sf { s_{n} = \frac{n}{2} (2a + (n-1)d) }$

» $\sf{ s_{40} = \frac{40}{2}(12 + 39(6)) }$

» $\sf { s_{40} = 20(12 + 234) }$

» $\sf { s_{40} = 20(246)}$

» Sum = 4920

Answer 2

Answer:

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