Arithmetic Sequences
An arithmetic sequence adds the same common difference d between consecutive terms, which is exactly what lets you jump straight to any term — or add up a whole batch of them — without writing out the list.
By the end you'll be able to find any term with \(a_n=a_1+(n-1)d\) and add up the first n terms with \(S_n = n(a_1+a_n)/2\).
Predict: the 10th term when \(a_1=3\) and \(d=4\). Set the sliders below to \(a_1=3\), \(d=4\), \(n=10\) and check — you should land on 39.
Drag a₁ and d to reshape the sequence; drag n to move the highlighted term. Each dot sits at its own value on the number line, one d step apart from the next. Below, the bars for the first n terms trace out a trapezoid — its area is exactly the partial sum \(S_n\). Hover or focus any dot or bar for exact readouts.
a₅ = a₁ + (5−1)d = 3 + 2(4) = 11 · S₅ = 5(3+11)/2 = 35
Every term in an arithmetic sequence is the last one plus the same fixed step d, so both "which term is this" and "what do the first n terms add up to" turn into short formulas instead of long lists.
Picture the dots on the number line above: start at \(a_1\) and hop by d each time. Ten hops from \(a_1\) lands you on \(a_{11}\), not \(a_{10}\) — the first term already sits at "hop 0," so the nth term is only \((n-1)\) hops from the start: \(a_n = a_1+(n-1)d\). To add up a run of terms, pair the first with the last, the second with the second-to-last, and so on — every pair lands on the same total, which is exactly what the trapezoid picture above is showing you.
Because \(a_{k+1}-a_k=d\) for every k, the nth term is \(a_n=a_1+(n-1)d\) — a closed formula that skips straight to any term without listing the ones before it. Pairing the first and last, second and second-to-last, etc. shows every pair sums to \(a_1+a_n\); with n terms making up n/2 such pairs (rounding gracefully for odd n), the sum of the first n terms is \(S_n = \dfrac{n(a_1+a_n)}{2}\), or equivalently \(S_n=\dfrac{n}{2}(2a_1+(n-1)d)\) after substituting the nth-term formula.
Stadium seating that adds the same number of seats each row, a savings plan that deposits the same fixed amount every month, a ladder's evenly spaced rungs — anywhere a quantity changes by a constant step, the total after n steps is a straight plug into \(S_n=n(a_1+a_n)/2\) instead of adding row by row or month by month.
Find \(a_{20}\) and \(S_{20}\) for 3, 7, 11, 15, ... . Here \(a_1=3\) and \(d=7-3=4\). \(a_{20}=a_1+(20-1)d=3+19(4)=3+76=\) 79. \(S_{20}=20(a_1+a_{20})/2=20(3+79)/2=20(82)/2=\) 820.
Same idea, different numbers: 1, 5, 9, 13, ... (\(a_1=1\), \(d=4\)). \(a_9=a_1+(9-1)d=1+8(4)=1+32=\) 33. Now finish it: \(S_9=9(a_1+a_9)/2=9(1+33)/2=9(34)/2=\) ____
Reveal the answer
\(S_9 = 9(34)/2 = 306/2 = \) 153. Set \(a_1=1\), \(d=4\), \(n=9\) on the sliders above and check the highlighted term and the partial-sum readout.
More info — why n is treated as the sequence's input
\(a_n=a_1+(n-1)d\) takes one input, the index n, and returns exactly one output, the term's value — the same "one input, one output" rule any function follows. Some textbooks even write it as \(f(n)\) instead of \(a_n\) to make that connection explicit. Reading the formula this way also explains why plugging in n itself (instead of n − 1) is such a common slip: it's evaluating the function at the wrong input. See the Khan Academy link in Dive deeper for more practice building the rule from a list of terms.
Check your understanding
For the arithmetic sequence 8, 13, 18, 23, ..., find \(a_{12}\).
The sum formula \(S_n = n(a_1+a_n)/2\) comes from pairing the first term with the last, the second with the second-to-last, and so on. What does each of those pairs add up to?
An arithmetic sequence's rule can be treated as a function of its index: \(f(n) = a_1 + (n-1)d\). For \(a_1=2\), \(d=3\), what is \(f(4)\)?
Find \(S_{10}\) for the arithmetic sequence with \(a_1=-3\), \(d=2\) (i.e. -3, -1, 1, 3, ...).
Recap
- An arithmetic sequence adds a fixed common difference d between consecutive terms.
- nth-term formula: \(a_n = a_1 + (n-1)d\) — the first term needs zero steps of d, so use \((n-1)\), not n.
- Partial-sum formula: \(S_n = n(a_1+a_n)/2\) — n terms times the average of the first and last, from pairing terms that all add to the same total.
- The nth term is a function of the index n: one n in, one term value out.
Dive deeper
- Khan Academy — Introduction to arithmetic sequences Build the nth-term rule from the common difference.
- Math is Fun — Arithmetic Sequences and Sums See where the S_n sum formula comes from.
Sources
- Arithmetic Sequences and Series