An arithmetic sequence adds the same number to get from each term to the next. That one fact gives a formula for the n-th term, and a pairing argument gives the sum of any number of terms without adding them up.
Core ideas
Two formulas, and one habit of checking. The habit is what keeps the formulas from being applied to sequences that do not qualify.
d = a₂ - a₁ is a guess until a₃ - a₂ and a₄ - a₃ agree with it. 2, 4, 8, 16 has first difference 2 and is not arithmetic; 1, 4, 9, 16 has differences 3, 5, 7 and is not either, though both look orderly.
Getting to a₁ takes no steps, to a₂ one step, to a₁₀ nine. The formula counts steps, and every off-by-one in this lesson is the difference between counting steps and counting terms.
The sequence 5, 9, 13, … is a list; S₃ = 27 is one number. Asking for "the 20th term" and "the sum of the first 20 terms" are different questions with different answers — 81 and 860 for the sequence in the worked example.
The material
The n-th term by counting steps, then the sum by writing it out twice.
A sequence in which the difference between consecutive terms is the same throughout. That number is the common difference d, and the recursive definition is aₙ = aₙ₋₁ + d together with a first term a₁.
3, 7, 11, 15, … has d = 4. 20, 17, 14, 11, … has d = -3; a negative common difference is ordinary and the formulas do not change. 5, 5, 5, 5, … has d = 0 and is arithmetic, which is a useful edge case rather than a trick.
Reaching aₙ from a₁ means adding d a certain number of times, and the number of times is the number of steps between the two positions:
If a₁ is the first term and d the common difference, then aₙ = a₁ + (n - 1)d for every n ≥ 1. Rearranged, d = (aₙ - a₁)/(n - 1), which recovers the common difference from the first term and any one other term. From two terms in general positions it is d = (aₙ - aₘ)/(n - m).
Plotted, an arithmetic sequence is a row of dots lying on a straight line of slope d — Lines, Functions and Graphs's slope, sampled at the integers. aₙ = a₁ + (n - 1)d is y = mx + c with the letters renamed, which is why a sequence with a constant difference is called linear.
A series is the sum of the terms of a sequence. The sum of the first n terms is the n-th partial sum Sₙ = a₁ + a₂ + … + aₙ, also written Sₙ = Σ aₖ with k running from 1 to n. It is a number, not a list.
Sₙ = n(a₁ + aₙ)/2. Substituting aₙ = a₁ + (n - 1)d gives the equivalent form Sₙ = (n/2)[2a₁ + (n - 1)d]. The first is convenient when the last term is known, the second when only a₁ and d are.
Write the sum out, and then write it out again in the opposite order: Sₙ = a₁ + (a₁ + d) + … + (aₙ - d) + aₙ and Sₙ = aₙ + (aₙ - d) + … + (a₁ + d) + a₁.
Add the two lines column by column. The first column gives a₁ + aₙ. The second gives (a₁ + d) + (aₙ - d) = a₁ + aₙ. Every column gives the same total, because each step of +d down the first line is matched by a step of -d up the second.
There are n columns and each totals a₁ + aₙ, so 2Sₙ = n(a₁ + aₙ), and dividing by 2 gives the formula. Nothing here depends on n being even or on the terms being positive.
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1 + 2 + 3 + … + 100 is arithmetic with a₁ = 1, d = 1, a₁₀₀ = 100. So S₁₀₀ = 100(1 + 100)/2 = 50(101) = 5050. The story is that Gauss found this at school by pairing 1 with 100, 2 with 99, and so on: fifty pairs, each totalling 101.
1, 3, 5, 7, … is arithmetic with a₁ = 1 and d = 2, so the n-th odd number is 1 + 2(n - 1) = 2n - 1 and Sₙ = n(1 + 2n - 1)/2 = n². The first five partial sums are 1, 4, 9, 16, 25, which is the fact that the odd numbers add up to the squares.
One last check that costs nothing: Sₙ = n(a₁ + aₙ)/2 says the sum is n times the average of the first and last terms. If 20 terms run from 5 to 81, the average is 43 and the sum should be near 860 — and it is exactly 860, because for an arithmetic sequence that average is exact rather than approximate.
Interactive
The nth term and the sum, each computed twice and compared
Method
Establish that it is arithmetic first. The formulas are false for anything else.
Compute a₂ - a₁, a₃ - a₂, a₄ - a₃. If they disagree, the sequence is not arithmetic and nothing below applies.
Both explicitly, before substituting anything. Most wrong answers in this topic come from the wrong d, and the commonest wrong d is the one with the sign flipped in a decreasing sequence.
For the sum, get aₙ first and then Sₙ = n(a₁ + aₙ)/2; or use Sₙ = (n/2)[2a₁ + (n - 1)d] and skip the intermediate term.
Sₙ should equal n times the midpoint of the first and last terms. A sum that is roughly double or half what that suggests is a dropped /2 or a doubled one.
Two questions that are often confused with each other, answered from the same two numbers.
The two forms of the sum formula are the same formula, and they agree here as they must. Which one to use is a question of what the problem gave you: a last term, or a common difference.
81 and 860 are the answers to two different questions. "The 20th term" is one term of the list; "the sum of the first 20 terms" adds all twenty. Reading the question twice is cheaper than computing the wrong one correctly.
For a faded arithmetic pass, use 18, 13, 8, 3, …. The supplied classification evidence is 13 - 18 = 8 - 13 = -5. Find a₁₅ and S₁₅, then check the sum as fifteen times the average of the first and fifteenth terms. Keep the negative difference visible in every substitution before opening the quiz.
Given any three of a₁, d, n, aₙ, Sₙ you should be able to produce the others, and to say which of the two sum formulas the given information calls for. The pairing argument should be reproducible from memory: it is three lines and it is the reason the formula is true.
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