How Do You Find the Sum of the First n Terms of an Arithmetic Sequence?

Use the arithmetic series formula

$$S_n=\frac{n}{2}\left(2u_1+(n-1)d\right),$$

where $u_1$ is the first term, $d$ is the common difference, and $n$ is the number of terms. If the final term $u_n$ is known, you can instead use .

This is part of SL 1.2, Sequences and Sigma Notation, which is studied by both Mathematics: Applications and Interpretation SL and HL students.

The formula works because an arithmetic series can be paired from opposite ends. The first and last terms sum to $u_1+u_n$, the second and second-last terms give the same total, and so on. Therefore, the mean term value is $\frac{u_1+u_n}{2}$. Multiplying this mean by gives . Since , substitution produces the other form.

Choosing the Formula

An arithmetic sequence has a constant difference between consecutive terms. First identify the information provided, then select the most convenient formula.

Worked Example

Find the sum of the first $20$ terms of

$$7,11,15,19,\ldots$$

The first term is $u_1=7$, and the common difference is

$$d=11-7=4.$$

Substitute into the formula:

$$S_{20}=\frac{20}{2}\left(2(7)+(20-1)(4)\right).$$

Therefore,

$$S_{20}=10(14+76)=900.$$

The sum of the first $20$ terms is $900$.

As a quick check, first calculate $u_{20}=7+19(4)=83$. The alternative formula then gives $S_{20}=\frac{20}{2}(7+83)=900$, confirming the result. This check is especially useful when the final term has already been found in an earlier part of a question. It also helps detect an incorrect common difference before the final answer.

A common misconception is using (nd) instead of $(n-1)d$. There are only $n-1$ increases from the first term to the $n$th term.

IB Exam Technique

For a calculate question, show the identified values of $u_1$, $d$, and $n$, followed by the correct substitution. Even when using a GDC, write the formula and expression before the numerical answer, keep full precision during calculations, and round only the final result if required.