The Boy Who Added a Hundred Numbers in Seconds
A schoolteacher once told a class to add every whole number from 1 to 100, expecting a quiet hour. One boy, Carl Friedrich Gauss, wrote a single number and sat back. He had spotted that the numbers pair up, and that idea is the whole engine behind the sum of n terms of an AP.
What Is the Sum of N Terms of an AP?
An arithmetic progression (AP) is a list of numbers where each term is the one before it plus a fixed step called the common difference, written $d$. The sum of its first $n$ terms, written $S_n$, is what you get when you add those terms together. Two formulas give it:
$$S_n = \frac{n}{2}\left[2a + (n-1)d\right]$$
$$S_n = \frac{n}{2}\left(a + l\right)$$
Here $a$ is the first term, $d$ the common difference, $n$ the number of terms, and $l$ the last term. Use the first form when you know $a$, $d$, and $n$; use the second when you already know the last term $l$. Both come from the same idea, and both describe the running total of an arithmetic progression.
How Do You Derive the Sum of N Terms of an AP?
The derivation is the pairing trick Gauss used, written in algebra. Start by writing the sum forwards:
$$S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d]$$
Now write the same sum backwards, last term first:
$$S_n = [a+(n-1)d] + [a+(n-2)d] + \cdots + a$$
Add the two lines term by term. Every vertical pair adds to the same value, $2a+(n-1)d$:
$$2S_n = [2a+(n-1)d] + [2a+(n-1)d] + \cdots + [2a+(n-1)d]$$
There are $n$ identical pairs, so:
$$2S_n = n\left[2a+(n-1)d\right]$$
Divide both sides by 2:
$$S_n = \frac{n}{2}\left[2a+(n-1)d\right]$$
That is the formula. Because the last term is $l = a+(n-1)d$, substituting it back gives the second form, $S_n = \frac{n}{2}(a+l)$. If you ever forget which is which, you can rebuild both from this one derivation, the same way you would rebuild the sum of arithmetic sequence result.
How Is It Linked to the Sum of the First N Natural Numbers?
The counting numbers $1, 2, 3, \ldots, n$ are themselves an AP with first term $a = 1$ and common difference $d = 1$. Drop those into the formula:
$$S_n = \frac{n}{2}\left[2(1) + (n-1)(1)\right]$$
$$S_n = \frac{n}{2}\left[n + 1\right]$$
$$S_n = \frac{n(n+1)}{2}$$
So the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$, and for $n = 100$ that is $\frac{100 \cdot 101}{2} = 5050$, exactly what Gauss wrote down. The general AP formula is just this idea stretched to any starting value and any step.
Where Does the Sum of an AP Actually Show Up?
"How much, in total, when the same amount keeps getting added?"
That question is everywhere, and the AP-sum formula answers it in one line instead of adding term by term.
Stacked and tapered objects. Logs in a pile, cans on a shelf, or seats in a theatre where each row holds a few more than the last are all APs, and the total count is $S_n$.
Steady raises and savings. A salary that climbs by a fixed amount each year, or a deposit that grows by the same step, gives a total earned or saved that is a direct AP sum.
Falling and stacking patterns in physics. Distances covered in equal time intervals under constant acceleration follow an AP, so total distance is $S_n$.
A springboard to other series. Once the AP sum is second nature, the same reverse-and-add reasoning motivates the sum of a GP, where the step multiplies instead of adds.
At Bhanzu, trainers introduce this formula by first making students add a long AP by hand until the tedium makes the shortcut feel earned, which is the moment the derivation sticks.
Examples of Sum of N Terms of an AP
The set runs from a direct substitution, through the most common trap, to finding an unknown number of terms and a real-world total.
Example 1
Find the sum of the first 20 terms of the AP with first term $a = 3$ and common difference $d = 5$.
Substitute $a = 3$, $d = 5$, $n = 20$:
$$S_{20} = \frac{20}{2}\left[2(3) + (20-1)(5)\right]$$
$$S_{20} = 10\left[6 + 95\right]$$
$$S_{20} = 10 \times 101$$
$$S_{20} = 1010$$
Final answer: $S_{20} = 1010$.
Example 2
Find the sum of the even numbers $2, 4, 6, \ldots, 100$.
Wrong attempt. A student sees the last number is 100 and writes $n = 100$, then computes $S_n = \frac{100}{2}[2(2) + (100-1)(2)] = 50[4 + 198] = 10100$. But there are not 100 even numbers up to 100. Plugging the last term in where the formula wants the number of terms inflates the answer.
Find $n$ first. The last term is $l = 100$, with $a = 2$ and $d = 2$:
$$100 = 2 + (n-1)(2)$$
$$98 = (n-1)(2)$$
$$n = 50$$
Correct sum. Now use $S_n = \frac{n}{2}(a+l)$:
$$S_{50} = \frac{50}{2}(2 + 100)$$
$$S_{50} = 25 \times 102$$
$$S_{50} = 2550$$
Final answer: $S_{50} = 2550$. Always separate "the last value" from "how many terms."
Example 3
An AP has first term $a = 7$, last term $l = 139$, and $45$ terms. Find the sum.
With the last term known, use the second form:
$$S_{45} = \frac{45}{2}(7 + 139)$$
$$S_{45} = \frac{45}{2}(146)$$
$$S_{45} = 45 \times 73$$
$$S_{45} = 3285$$
Final answer: $S_{45} = 3285$.
Example 4
How many terms of the AP $3, 7, 11, \ldots$ add up to $210$?
Here $a = 3$, $d = 4$, and $S_n = 210$. Set up the equation:
$$210 = \frac{n}{2}\left[6 + (n-1)(4)\right]$$
$$210 = \frac{n}{2}\left[4n + 2\right]$$
$$210 = n(2n + 1)$$
$$2n^2 + n - 210 = 0$$
Solve the quadratic, keeping only the positive whole-number root:
$$n = \frac{-1 + \sqrt{1 + 1680}}{4} = \frac{-1 + 41}{4} = 10$$
Final answer: $n = 10$ terms. The step students skip most often is checking that $n$ came out a whole number, because a fractional $n$ means the target sum is not reachable with this AP.
Example 5
Find the sum of the first 100 natural numbers.
The numbers $1$ to $100$ form an AP with $a = 1$, $d = 1$, $n = 100$:
$$S_{100} = \frac{100}{2}\left[2(1) + (100-1)(1)\right]$$
$$S_{100} = 50\left[2 + 99\right]$$
$$S_{100} = 50 \times 101$$
$$S_{100} = 5050$$
Final answer: $S_{100} = 5050$, matching the natural-numbers shortcut $\frac{n(n+1)}{2}$.
Example 6
A junior engineer starts at a salary of 30,000 a year, with a fixed raise of 1,500 each year. What is the total earned over the first 10 years?
The yearly salaries form an AP with $a = 30000$, $d = 1500$, $n = 10$:
$$S_{10} = \frac{10}{2}\left[2(30000) + (10-1)(1500)\right]$$
$$S_{10} = 5\left[60000 + 13500\right]$$
$$S_{10} = 5 \times 73500$$
$$S_{10} = 367500$$
Final answer: total earned is $367{,}500$ over the ten years.
Common Mistakes
Mistake 1: Treating the last term as the number of terms
Where it slips in: A series is given by its values, like $2, 4, \ldots, 100$, with no term count stated.
Don't do this: Read the final value $100$ as $n$ and substitute it straight into the sum formula.
The correct way: Find $n$ first from $l = a + (n-1)d$, then substitute. The last term and the term count are different numbers, and the formula wants the count. Learners meeting the formula for the first time make this substitution slip more than any other.
Mistake 2: Computing the common difference wrong
Where it slips in: Identifying $d$ from the first few terms.
Don't do this: Take $d$ to be the second term, or subtract in the wrong order and lose a sign on a decreasing AP.
The correct way: The common difference is always later term minus earlier term, $d = a_2 - a_1$. For $10, 7, 4, \ldots$ that is $d = 7 - 10 = -3$, and a negative $d$ must stay negative all the way through the sum.
Mistake 3: Forgetting to reject a non-whole n
Where it slips in: Solving for the number of terms from a target sum, which produces a quadratic.
Don't do this: Accept both roots, including a negative or fractional one, as valid answers.
The correct way: The number of terms is a positive whole number, so discard any root that is negative or non-integer. The second-guesser who solves the quadratic correctly but keeps $n = -14$ alongside $n = 10$ needs only to remember what $n$ physically counts.
Conclusion
The sum of n terms of an AP is $S_n = \frac{n}{2}[2a+(n-1)d]$, with the shorter $S_n = \frac{n}{2}(a+l)$ used when the last term is known.
The formula comes from adding the series to its own reverse, so every pair contributes the same $2a+(n-1)d$.
The counting numbers are an AP with $a=1$, $d=1$, giving the special case $\frac{n(n+1)}{2}$.
The most frequent error is confusing the last term with the number of terms, so solve for $n$ before summing.
AP sums count stacked objects, steady raises, and equal-interval distances in one step instead of term by term.
To work through more arithmetic-series problems with a teacher, explore Bhanzu's algebra tutor sessions, a dedicated high school math tutor, or live math classes online. Want a trainer to walk your child through APs and the Class 10 sequences chapter? Book a free demo class.
Read More
Arithmetic Sequence Formula — the term-by-term view of the same progression.
Sequences in Algebra — how arithmetic, geometric, and other patterns are defined.
Sum of Cubes of N Natural Numbers — a related closed-form sum.
Nth Term of a GP — the geometric counterpart to the nth term of an AP.
Sum of an Infinite GP — what happens when a series never ends.
Sum of Natural Numbers Formula — the $\frac{n(n+1)}{2}$ result on its own.
Was this article helpful?
Your feedback helps us write better content