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

Factors and multiples
ANSWERED To determine the sum of all multiples of 3 between 1 and 1000 2021-08-05
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 $$\displaystyle{a}_{{{1}}}={3}$$ and the common difference is $$\displaystyle{d}={3}$$.
The n-th term of the arithmetic sequence is
$$\displaystyle{a}_{{{n}}}={3}+{\left({n}-{1}\right)}{3}$$
$$\displaystyle{a}_{{{n}}}={3}+{3}{n}-{3}$$
$$\displaystyle{a}_{{{n}}}={3}{n}$$
Therefore the n-th term of the arithmetic sequence is $$\displaystyle{a}_{{{n}}}={3}{n}$$.
When $$\displaystyle{a}_{{{n}}}={999}$$,
$$\displaystyle{999}={3}{n}$$
$$\displaystyle{\frac{{{999}}}{{{3}}}}={n}$$
$$\displaystyle{333}={n}$$
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
$$\displaystyle{S}_{{{n}}}={\frac{{{n}}}{{{2}}}}{\left({a}+{L}\right)}$$
where $$\displaystyle{S}_{{{n}}}$$ is the sum of n terms and L is the last term of the arithmetic sequence.
For the given arithmetic sequence $$\displaystyle{a}={3},{L}={999}$$ and $$\displaystyle{n}={333}$$,
$$\displaystyle{S}_{{{333}}}={\frac{{{333}}}{{{2}}}}{\left({3}+{999}\right)}$$
$$\displaystyle{S}_{{{333}}}={\frac{{{333}}}{{{2}}}}\times{1002}$$
$$\displaystyle{S}_{{{333}}}={333}\times{501}$$
$$\displaystyle{S}_{{{333}}}={166833}$$
Therefore the sum of all multiples of 3 between 1 and 1000 is 166833.