Sum of Arithmetic Sequence Formula — Proof, Examples

TL;DR

The sum of arithmetic sequence formula adds the first $n$ terms of an arithmetic progression: $S_n = \frac{n}{2}\big(2a + (n-1)d\big) = \frac{n}{2}(a + l)$. This article derives both forms by the pairing trick, defines every variable, works through six examples from one-step sums to finding $n$ from a known total, and clears up the mistakes that cost the most marks.

What Is the Sum of an Arithmetic Sequence?

An arithmetic sequence is an ordered list of numbers with a constant gap between consecutive terms, called the common difference $d$. The sequence $3, 7, 11, 15, \ldots$ has first term $a = 3$ and $d = 4$. Adding its terms gives an arithmetic series, and the sum of the first $n$ terms is written $S_n$.

The two forms of the formula use different inputs:

$$\boxed{;S_n = \frac{n}{2}\big(2a + (n-1)d\big);}\qquad\boxed{;S_n = \frac{n}{2}(a + l);}$$

Symbol	Meaning	Notes
$S_n$	Sum of the first $n$ terms	The quantity you want
$n$	Number of terms being added	Always a positive integer
$a$	First term ($a_1$)	Where the sequence starts
$d$	Common difference	The constant step between terms
$l$	Last term ($a_n$)	Equals $a + (n-1)d$

Use the first form when you know $a$, $d$, and $n$. Use the second when you already know the first and last terms — it is just the average of the endpoints, $\frac{a+l}{2}$, multiplied by how many terms there are.

How Do You Derive the Sum Formula?

The derivation is the pairing trick, and it is worth seeing once rather than memorising the result blind. Write the series forwards, then write it again backwards underneath:

$$S_n = a + (a+d) + (a+2d) + \ldots + l$$ $$S_n = l + (l-d) + (l-2d) + \ldots + a$$

Add the two lines column by column. Every column sums to $a + l$ — the increases on top cancel the decreases on the bottom. There are $n$ columns, so:

$$2S_n = n(a + l).$$

Divide by $2$:

$$S_n = \frac{n}{2}(a + l).$$

Now substitute the last term $l = a + (n-1)d$ to reach the other form:

$$S_n = \frac{n}{2}\big(a + a + (n-1)d\big) = \frac{n}{2}\big(2a + (n-1)d\big).$$

Both forms are the same statement. The pairing argument is the one a child can follow; it is also the same idea behind the sum of natural numbers formula, which is just this formula with $a = 1$ and $d = 1$.

Examples of the Sum of Arithmetic Sequence Formula

Example 1

Find the sum of the first 10 terms of $3, 7, 11, 15, \ldots$

Here $a = 3$, $d = 4$, $n = 10$. Use the first form:

$$S_{10} = \frac{10}{2}\big(2(3) + (10-1)(4)\big) = 5(6 + 36) = 5 \times 42 = 210.$$

Final answer: $210$.

Example 2

Add the even numbers $2 + 4 + 6 + \ldots + 100$.

Wrong attempt. A student reasons "the last term is $100$, so $n = 100$" and writes $S = \frac{100}{2}(2 + 100) = 50 \times 102 = 5100$. Take a step back: there are only $50$ even numbers between $2$ and $100$, not $100$ of them. The count of terms is not the value of the last term.

Correct. First find $n$ from $l = a + (n-1)d$ with $a = 2$, $d = 2$, $l = 100$:

$$100 = 2 + (n-1)(2) ;\Rightarrow; 98 = 2(n-1) ;\Rightarrow; n = 50.$$

Then $S_{50} = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550$.

Final answer: $2550$.

Example 3

The first term is $10$, the last term is $40$, and there are $7$ terms. Find the sum.

Both endpoints are known, so the $(a + l)$ form is fastest:

$$S_7 = \frac{7}{2}(10 + 40) = \frac{7}{2}(50) = 175.$$

Final answer: $175$.

Example 4

Find $5 + 8 + 11 + \ldots$ up to $20$ terms.

$a = 5$, $d = 3$, $n = 20$:

$$S_{20} = \frac{20}{2}\big(2(5) + (20-1)(3)\big) = 10(10 + 57) = 10 \times 67 = 670.$$

Final answer: $670$.

Example 5

For what value of $n$ does $2 + 5 + 8 + \ldots = 155$?

Set up the first form with $a = 2$, $d = 3$:

$$\frac{n}{2}\big(2(2) + (n-1)(3)\big) = 155 ;\Rightarrow; \frac{n}{2}(3n + 1) = 155.$$

Multiply out: $3n^2 + n - 310 = 0$. The quadratic formula gives

$$n = \frac{-1 \pm \sqrt{1 + 3720}}{6} = \frac{-1 \pm 61}{6}.$$

The positive root is $n = 10$.

Final answer: $n = 10$. Check: $\frac{10}{2}(4 + 27) = 5 \times 31 = 155$. ✓

Example 6

A stadium has $15$ seats in the front row and $2$ more seats in each row behind it. How many seats are there in $30$ rows?

This is an arithmetic series with $a = 15$, $d = 2$, $n = 30$:

$$S_{30} = \frac{30}{2}\big(2(15) + (30-1)(2)\big) = 15(30 + 58) = 15 \times 88 = 1320.$$

Final answer: $1320$ seats.

Why the Sum Formula Matters — Beyond Adding Lists

Arithmetic series were studied because totals of steadily-changing quantities show up everywhere, and adding term by term is hopeless past a handful of terms.

Stacked objects. Logs in a triangular pile, cans in a sloped display, seats fanning out in an auditorium — each row adds a fixed amount, so the total is an arithmetic series.
Loan and savings schedules. A fixed-increment repayment plan (pay a little more each month) totals to an arithmetic sum; the formula tells you the full amount repaid without a spreadsheet.
Physics of constant acceleration. Distance covered in successive equal time intervals under constant acceleration grows arithmetically — Galileo's odd-number rule ($1, 3, 5, 7, \ldots$) is an arithmetic series, and its partial sums are the perfect squares.
The bridge to calculus. Partial sums of sequences are the discrete ancestor of the integral. Before $\int$ existed, mathematicians summed series like this one to approximate areas.

The destination, a few years down the line, is the same averaging idea: the sum is "how many terms" times "the average term," and for an arithmetic sequence the average term is exactly the midpoint of the first and last.

Tripping Points to Avoid

Mistake 1: Confusing the last term with the number of terms

Where it slips in: Sums written by their endpoints, like "$3 + 6 + 9 + \ldots + 90$."

Don't do this: Take $n = 90$ because $90$ is the last term.

The correct way: Solve $l = a + (n-1)d$ for $n$ first. For $3 + 6 + \ldots + 90$ with $a = 3$, $d = 3$: $90 = 3 + (n-1)(3)$ gives $n = 30$.

Mistake 2: Using the wrong form for the information you have

Where it slips in: Problems that give the last term directly but no common difference.

Don't do this: Reach for $\frac{n}{2}(2a + (n-1)d)$ and try to back out $d$ when the last term is already sitting in front of you.

The correct way: If you know $a$, $l$, and $n$, use $S_n = \frac{n}{2}(a + l)$ — no $d$ needed. Match the form to your inputs.

Mistake 3: Forgetting the formula needs a constant common difference

Where it slips in: Sequences that look regular but are not arithmetic, such as $2, 4, 8, 16$ (that one is geometric).

Don't do this: Apply the arithmetic sum formula to a sequence whose gaps change.

The correct way: Check that consecutive differences are equal before using the formula. If the gap is constant, it is arithmetic; if the ratio is constant, it is geometric and needs a different formula.

Mistake 4: Sign errors with a negative common difference

Where it slips in: Decreasing sequences like $20, 17, 14, \ldots$ where $d = -3$.

Don't do this: Drop the minus sign on $d$ when substituting into $(n-1)d$.

Conclusion

The sum of arithmetic sequence formula is $S_n = \frac{n}{2}\big(2a + (n-1)d\big) = \frac{n}{2}(a + l)$, two equivalent forms for the total of the first $n$ terms.
The pairing derivation — write the series forwards and backwards, add, divide by two — is short enough to reconstruct any time you forget the result.
Choose the $(2a + (n-1)d)$ form when you know the common difference, and the $(a + l)$ form when you know the last term.
The most common mistake is reading the last term as the number of terms; always solve for $n$ first.
It is the discrete cousin of the integral and the direct generalisation of the sum-of-naturals formula.

Practice These Three Before Moving On

Find the sum of the first $25$ terms of $4, 9, 14, 19, \ldots$
Add the multiples of $5$ from $5$ to $200$ (find $n$ first).
For what $n$ does $1 + 4 + 7 + \ldots = 145$?

Answer to Question 1: $1450$. Work through Questions 2 and 3 the same way; if you stumble on finding $n$, return to Mistake 1.