Arithmetic Sequences: Constant Difference, Explained

An arithmetic sequence adds the same number at every step. That number is the common difference, usually written d, and it is the only thing you need in order to extend the sequence, jump to any term without listing the ones before it, or add a whole run of terms at once. The University of Waterloo's CEMC sequences and series toolkit states the defining property plainly: the difference between consecutive terms is constant.

This is the narrowest and most common of the families catalogued in the guide to number sequence puzzles. It is worth doing properly, because the two formulas involved are usually handed over as things to memorise when both of them can be reconstructed from scratch in about ten seconds.

Recognising one: subtract, and see if it settles

The test is one subtraction per gap. Take 4, 11, 18, 25, 32. The gaps are 7, 7, 7, 7. A constant first-difference row is the definition, so this sequence is arithmetic with a = 4 and d = 7.

If that row is not constant, stop treating the sequence as arithmetic. The next two tests are dividing consecutive terms, which detects a constant ratio, and taking differences a second time, which detects a quadratic rule. NRICH's article on divided differences by T. W. Körner sets out the general version of this: repeatedly differencing a polynomial of degree m produces a row of zeros after at most m + 1 passes, so an arithmetic sequence is simply the degree-one case where the first difference row is already flat.

Deriving the nth-term rule instead of memorising it

You want term number n without writing out everything before it. Count the steps.

• Term 1 is a. Zero steps taken.
• Term 2 is a + d. One step.
• Term 3 is a + 2d. Two steps.
• Term n is a + (n − 1)d. n − 1 steps.

That is the whole derivation. The n − 1 is not a quirk of notation; it is the count of gaps between the first term and the nth, and there is always one fewer gap than there are terms.

Apply it to 4, 11, 18, 25, 32. The rule is 4 + 7(n − 1), which expands to 7n − 3. Check both ends before trusting it: at n = 1, 7 − 3 = 4, correct; at n = 5, 35 − 3 = 32, correct. The twentieth term is 7 × 20 − 3 = 137, and you never had to write out terms six to nineteen.

The expanded form, 7n − 3, is worth keeping. It shows the common difference sitting where the gradient of a straight line would sit, and the constant −3 as the value the line would take at position zero.

The sum is pairing, not a formula

The sum of the first n terms of an arithmetic sequence is

S = n/2 × (first term + last term)

which is the number of terms multiplied by the average of the two ends. The reason is worth seeing once, because after that the formula is impossible to forget.

Write the run forwards, then write it backwards underneath, and add the two lines column by column. Every column contains one term from the front and its partner the same distance from the back, and every such pair totals the same amount. You have now added the sequence to itself, so halve the result.

MacTutor's biography of Carl Friedrich Gauss records the well-known schoolroom story of exactly this move: his teacher Büttner and his assistant Martin Bartels were reported to be amazed when Gauss summed the integers from 1 to 100 instantly by spotting that the sum was fifty pairs each totalling 101 — giving 5,050. The anecdote is a staple of mathematical folklore and the details vary between retellings, but the method it describes is sound and is the standard derivation of the formula.

For our running example, the sum of the first twenty terms is 20/2 × (4 + 137) = 10 × 141 = 1,410. The alternative form, n/2 × (2a + (n − 1)d), gives 10 × (8 + 133) = 10 × 141, which agrees. Running a second form as a check costs one line and catches a slipped term count.