What is the sum of the multiples of 6 between 6 and 999?
here's what i did:
n = 996/6 = 166?
d = 6
t1 = 6

sn = n/2 [2 (t1) + (n-1)d]

Your method is completely correct. The first term will be 6 and each subsequent term can be obtained by adding 6 to the previous one, meaning the common difference is 6. The number of terms is given by the highest number that is divisible by 6 and dividing it by 6; that is 996/6 = 166
Then we simply apply the formula for arithmetic sequence sum:
S = n/2 [2a₁ + (n - 1)d]
S = 166/2 [ 2(6) + (166 - 1)6]
S = 83,166

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Definitely