Question

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

Factors and multiples
ANSWERED
asked 2021-08-04
To determine the sum of all multiples of 3 between 1 and 1000

Answers (1)

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.
0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...