A constant step between terms turns counting into algebra — and Gauss's pairing trick turns a thousand additions into one formula.
9 min read4 worked examples
The nth term of an arithmetic sequence
An arithmetic sequence is a sequence in which each term differs from the previous one by a fixed amount, the common difference $d$. Starting from a first term $a$, the sequence runs $a,\ a+d,\ a+2d,\ a+3d,\dots$, so the $n$th term is
$u_n = a+(n-1)d$
Each new term needs one more "step" of size $d$ than the last, and the $(n-1)$ counts how many steps have been taken to reach term $n$ from the first term. If $d>0$ the sequence is increasing; if $d<0$ it is decreasing; if $d=0$ every term is identical.
The sum of an arithmetic series
An arithmetic series is the sum of the terms of an arithmetic sequence. Adding $n$ terms one at a time is slow — instead, there's a closed-form formula for the sum of the first $n$ terms, $S_n$:
where $l=u_n=a+(n-1)d$ is the last (nth) term of the series. The two forms are identical — the second simply substitutes $l$ in place of $a+(n-1)d$ — but the $\frac{n}{2}(a+l)$ form is often faster when the last term is already known.
ProofThe sum formula, by pairing forwards and backwards
1
Write $S_n$ as the sum of terms from first to last: $S_n = a+(a+d)+(a+2d)+\cdots+(a+(n-1)d)$.
Direct definition of the sum of $n$ terms.
2
Write the same sum again, but listed backwards, from last to first: $S_n = (a+(n-1)d)+(a+(n-2)d)+\cdots+a$.
Reversing the order doesn't change the value of the sum.
3
Add the two versions of $S_n$ together, term by term (1st with 1st, 2nd with 2nd, and so on): each pair sums to $\big[a+(a+(n-1)d)\big] = 2a+(n-1)d$.
In every paired column, the $d$-multiples are complementary and cancel to the same total.
4
There are $n$ such pairs (one per term), so $2S_n = n\big[2a+(n-1)d\big]$.
Adding two copies of $S_n$ gives $2S_n$, and each of the $n$ columns contributes the same constant sum.
5
Divide both sides by 2: $S_n = \dfrac{n}{2}\big[2a+(n-1)d\big]$.
Isolates $S_n$, giving the standard sum formula.
∎ proven
The special case: summing the natural numbers
Taking $a=1$ and $d=1$ gives the arithmetic sequence of natural numbers $1,2,3,\dots,n$, so the sum formula collapses to the famous result $\displaystyle\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$. Legend has it a young Gauss spotted exactly this pairing trick — adding $1+2+\cdots+100$ forwards and backwards — to sum the numbers 1 to 100 almost instantly.
Worked example 1
Finding a term from the nth-term formula
An arithmetic sequence has first term $a=7$ and common difference $d=-3$. Find the 15th term.
MethodSubstitute directly into $u_n=a+(n-1)d$
1
$u_{15} = a+(15-1)d = 7+14(-3)$.
Substitute $n=15$, $a=7$, $d=-3$ into the nth-term formula.
2
$u_{15} = 7-42 = -35$.
Evaluate.
▪ solved
Examiner's eye
Since $d<0$ here, the sequence is decreasing — a quick sanity check that a large negative 15th term is sensible.
Worked example 2
Finding a and d from two given terms
The 3rd term of an arithmetic sequence is 11 and the 8th term is 31. Find $a$ and $d$, and hence find $S_{20}$.
MethodForm simultaneous equations from the nth-term formula, then apply the sum formula
1
$u_3 = a+2d = 11$ and $u_8 = a+7d = 31$.
Substitute $n=3$ and $n=8$ into $u_n=a+(n-1)d$.
2
Subtract the first equation from the second: $5d = 20 \Rightarrow d=4$.
Substitute $n=20$, $a=3$, $d=4$ into the sum formula.
▪ solved
Strategy
Whenever two terms are given, subtracting one nth-term equation from another eliminates $a$ immediately — this is the fastest route to $d$, after which $a$ falls out by back-substitution.
Worked example 3
Applying the sum formula to a real-world context
Bricks are stacked in a pile with 15 rows. The top row has 3 bricks, and each row below has 2 more bricks than the row above it. Find the total number of bricks in the pile.
MethodIdentify a, d and n from the context, then apply the sum formula
1
The row sizes form an arithmetic sequence with first term $a=3$ (bricks in the top row) and common difference $d=2$ (each row down has 2 more bricks), with $n=15$ rows in total.
Translate the physical description into the three values the sum formula needs.
2
$u_{15} = a+(15-1)d = 3+14(2) = 31$.
Find the last term (the bottom row) first, since $S_n=\frac{n}{2}(a+l)$ needs it.
Substitute the first and last terms into the sum formula.
4
The pile contains 255 bricks.
Restate the final answer in the context of the original question.
▪ solved
Translating a context into a, d, n
The arithmetic here is routine — the real skill is recognising that "each row has 2 more than the one above" describes a common difference, and that a fixed number of rows is $n$. Once $a$, $d$ and $n$ have been identified, a context question becomes exactly the same calculation as an abstract one.
Worked example 4
Solving for n given a target sum · Challenge
An arithmetic series has first term $a=5$ and common difference $d=2$. Find the least value of $n$ for which $S_n > 500$.
MethodForm a quadratic inequality in $n$, then solve and interpret the result
Using the quadratic formula: $n = \dfrac{-4\pm\sqrt{16+2000}}{2} = \dfrac{-4\pm\sqrt{2016}}{2}$.
The quadratic $n^2+4n-500$ doesn't factorise neatly, so solve $n^2+4n-500=0$ to find the boundary.
4
$\sqrt{2016}\approx44.9$, so $n\approx\dfrac{-4+44.9}{2}\approx20.45$ (rejecting the negative root, since $n$ must be a positive integer).
Only the positive root is meaningful for a term number.
5
Since $n^2+4n-500$ is increasing for $n>0$, the inequality holds once $n$ exceeds $\approx20.45$; testing integers confirms $n=21$ gives $S_{21}=21^2+4(21)=441+84=525>500$, while $n=20$ gives $S_{20}=400+80=480<500$.
Since $n$ must be a whole number of terms, round up from the exact boundary and verify with the original formula.
6
The least value of $n$ is $21$.
Final answer.
▪ solved
Look deeper — always verify after rounding
Solving the quadratic gives a real-number boundary, but $n$ must be a positive integer — so the exact root is never the final answer on its own. Always test the integers either side of the boundary in the original sum formula, exactly as in step 5, to confirm which one first satisfies the inequality.
Common mistake
Confusing $n$ (the number of terms) with $d$ (the common difference), or forgetting that the nth-term formula uses $(n-1)$ steps, not $n$ steps, from the first term — writing $u_n=a+nd$ is a common slip that shifts every term by one position. Always check with $n=1$: the formula must return $u_1=a$ exactly.
Check yourself
An arithmetic sequence has $u_1=4$ and $u_{10}=40$. Find the common difference $d$, and the sum of the first 25 terms. (Hint: use the two given terms to find $d$ first, then apply the sum formula.)