# Math

The sum of multiples of 6 between 6 & 999?

I got 83166, but my answer key says 82665.. Is my answer wrong? I used the term formula to find 166 terms multiples of 6, and then the sum formula to get 83166.

1. So you have 6 + 12 + 18 + ... + 996
this is an AS where a = 6, d = 6
so 996 = 6 + (n-1)(6)
996 = 6 + 6n - 6
6n = 996
n = 166 , you are right so far

sum(166) = 83(first + last)
= 83(6 + 996) = 83166 , you had that too

or using the more standard sum formula
S(n) = (n/2)(2a + (n-1)d)
= (166/2)(12 + 165(6) )
= 83166

The only thing I see is they said "between",
so exclude the first term but keep last term
83166 - 6 = 83160

check on that, now we only have 165 terms
sum = (165/2)(12 + 996) = 83160 , as above

posted by Reiny

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