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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What is the sum of the multiples of 6 between 6 and 999?Here's what I did:n = 996/6 = 166?d = 6t1 = 6Sn = n/2 [2 (t1) + (n-1)d]

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Related Questions

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]