Question

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]

Answer 1

the series is 12 + 8 + 24 + 30 + ... + 996
so, the frist term is 12

Answer 2

looks u were right :)
then use the formula above

Answer 3

I did but it didn't work. The correct answer is 82, 665 but my answer was 83, 166.

Answer 4

isn't the series supposed to be 6, 12, 18.....?

Answer 5

82.665

Answer 6

btw, there is a bit typo from me
the series should is 12 + 18 + 24 + 30 + ... + 996

T1 = 12 (because the word "between 6 and 999" means that numbers are more than 6 and less than 999)

d = 6

to get the value of n, u can find out by using the formula :
Tn = T1 + (n-1)d
996 = 12 + (n-1)6
996 = 12 + 6n - 6
996 = 6n + 6
996 - 6 = 6n
990 = 6n
n = 990/6 = 165
so, there are 165 terms

thus, the sum all terms becomes
Sn = n/2 [2 (t1) + (n-1)d]
S165 =165/2 (2*12 + (165-1)6)
S165 =165/2 (24 + 164*6)
S165 =165/2 (24 + 984)
S165 =165/2 (1008)
S165 =83,160

Answer 7

Let us put a Little logic in things. The solution cannot be 82.665 as a sum of multiples of 6 is also a multiple of 6 and an odd number cannot be multiple of 6. Therefore, the solution 82.665 is discarded. You have got the right solution :83.166

Answer 8

@CarlosGP , do u know the means of "between 6 and 999" ?
actually, it is