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

The first 40 positive integers that are divisible by 6 are 6, 12, 18, 24…
a = 6 and d = 6. We need to find S₄₀
Sₙ = n/2[2a + (n-1)d]
S₄₀ = 40/2[2(6) + (40-1)6] = 20[12 + (39)6] = 20(12 + 234) = 20 × 246 = 4920