Sum of Multiples in a Range

The multiples of a number form an arithmetic progression. To sum them in a range: find the first and last terms, work out how many there are, then apply $S_n = \frac{n}{2}(a + l)$.

Find the sum of all the multiples of $3$ between $1$ and $100$. [4]

The working

Step 1, identify the AP. The multiples of $3$ are $3, 6, 9, \ldots$, so $a = 3, \qquad d = 3.$ The largest multiple of $3$ below $100$ is $99$, so the last term is $l = 99$.

Step 2, find the number of terms using $l = a + (n-1)d$: $99 = 3 + (n-1)3 \implies 96 = 3(n-1) \implies n - 1 = 32 \implies n = 33 \quad \text{(M1, A1)}$

Step 3, apply the sum formula $S_n = \frac{n}{2}(a + l)$: $S_{33} = \frac{33}{2}(3 + 99) = \frac{33}{2}(102) = 33 \times 51 \quad \text{(M1)}$

Step 4, evaluate: $S_{33} = 1683 \quad \text{(A1)}$

Where the marks are won and lost

• The last term is $99$, not $100$: $100$ is not a multiple of $3$. Getting $l$ right is what fixes $n$.
• Count carefully: $n = 33$, from $l = a + (n-1)d$. An off-by-one here throws the whole sum.
• $S_n = \frac{n}{2}(a + l)$ is quickest when you know the last term; $\frac{n}{2}(2a + (n-1)d)$ gives the same answer.

Common mistakes

• Using $l = 100$ or $n = 34$.
• Forgetting to divide by $2$ in the sum formula.
• Summing the integers $1$ to $100$ rather than only the multiples of $3$.

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