To determine the sum of all multiples of 3 between 1 and 1000

Question

Nola Robson · Accepted Answer

The multiple of 3 are 3, 6, 9, 12 ..., 999.
The sequence of multiple of 3 are in an arithmetic sequence with the first term is

a_{1} = 3

and the common difference is

d = 3

.
The n-th term of the arithmetic sequence is

a_{n} = 3 + (n - 1) 3

a_{n} = 3 + 3 n - 3

a_{n} = 3 n

Therefore the n-th term of the arithmetic sequence is

a_{n} = 3 n

.
When

a_{n} = 999

,

999 = 3 n

\frac{999}{3} = n

333 = n

Therefore the 333-th term of the arithmetic sequence is 999.
The sum of n terms of an arithmetic sequence is given by the formula

S_{n} = \frac{n}{2} (a + L)

where

S_{n}

is the sum of n terms and L is the last term of the arithmetic sequence.
For the given arithmetic sequence

a = 3, L = 999

and

n = 333

,

S_{333} = \frac{333}{2} (3 + 999)

S_{333} = \frac{333}{2} \times 1002

S_{333} = 333 \times 501

S_{333} = 166833

Therefore the sum of all multiples of 3 between 1 and 1000 is 166833.

To determine the sum of all multiples of 3 between 1 and 1000.