Common difference, and why the nth term is a + (n − 1)d

0:00Here is a pattern that grows in stages. Stage one is a single tile. Stage two adds four tiles, one on each arm. Five in all. Stage three adds four more. Nine. Stage four, thirteen. One, five, nine, thirteen, and it carries on seventeen, twenty-one. The gap between one stage and the next is four, every single time. So we could get to stage twenty by adding four over and over. But we would rather not.

0:35Instead of counting tiles, count additions. Stage one is one, with nothing added to it. Stage two is one plus four. Stage three is one plus four plus four. Stage four is one plus four plus four plus four. Now collect the repeats, because four added three times is three fours. One plus nothing. One plus one four. One plus two fours. One plus three fours. And there is the number we came for, sitting in front of each four.

1:11Look at the multipliers on their own: nothing, one, two, three. Against the stage numbers one, two, three, four. Each multiplier is one less than its stage, and that is not a coincidence to be memorised. It is a count of how many times you added four to get there. Stand on stage one. How many additions have you performed to be standing there? None. You started there. Stage two took one addition, stage three took two, stage four took three.

1:45There is always one fewer addition than there are stages, because the first stage is where the counting starts, not where it moves. So the rule is one, plus n minus one, lots of four. Which tidies to four n minus three. Stage twenty, without building nineteen stages first: seventy-seven. It is worth seeing what goes wrong if you forget the minus one. Suppose the rule were one plus n fours, with no adjustment.

2:18Put in one. That gives one plus four, which is five. But stage one is a single tile. The formula has walked one step too far before you asked it to move at all. And it stays exactly one step too far, forever: at every position it overshoots by four. There is only one situation where the two agree, and it is the boring one. If the step is nothing, then taking one step too many costs you nothing, and both formulas are right.

2:52For every other step, one of them is wrong at every single position. Now write it with letters, because the tiles were never the point. Call the first term a, and call the fixed gap d, the common difference. Then the terms run a, a plus d, a plus two d, a plus three d. Number the additions underneath: nothing, one, two, three. The nth term is a plus n minus one, times d.

3:22A sequence like this is called an arithmetic progression, and those two letters are the whole of it. Give me a and d and I can hand you any term you like, at any position, without building the ones before it. Two numbers, and an infinite list is decided. There is a second way to say the same thing, and you have met it before. Start at a. Then each term is the term before it, plus d, from the second term onwards.

3:54For the tiles: start at one, then add four each time. Both descriptions produce exactly the same list. They cost very different amounts to use. Reaching the four hundredth term by adding takes three hundred and ninety-nine additions. Reaching it from the formula takes two operations, whatever the position is. Which is the whole reason for doing the algebra: the step rule tells you how the list is built, and the formula lets you skip to anywhere in it.

4:29Nothing in any of that required the list to climb. Take eleven, seven, three, minus one, minus five. The gap is the same every time: minus four. That is an arithmetic progression, with a first term of eleven and a common difference of minus four. It is not a different kind of object that happens to go downwards. It is the same object with a negative d. Its rule comes out as fifteen minus four n, and if you prefer, that is eleven plus n minus one, times minus four.

5:07The sign of the common difference is the only thing deciding whether a progression climbs or falls. Nor does anything require whole numbers. Here is one: a half, five halves, nine halves, thirteen halves. The gap is two, and the first term is a half, so the nth term is four n minus three, all over two. And here is another: one point five, three point five, five point five, seven point five.

5:39The gap is two again, and the first term is one point five, so the nth term is two n minus a half. Two progressions with exactly the same common difference, and they are not the same progression. Run them both as far as you like and they never once land on the same number. That is what it means to say d names a family and a picks one out of it. You need both.

6:08Which of the lists we have met are arithmetic progressions? There is exactly one test, and it is not whether there is a formula. Take the gaps between consecutive terms and see whether they are all the same. The counting numbers, gap one. The odd numbers, gap two. The tiles, gap four. One, four, seven, ten, gap three. Four progressions. The triangular numbers, the square numbers, the sequence built from the two before it, and the unit fractions. Four that are not.

6:43The square numbers are the instructive failure. One, four, nine, sixteen, twenty-five, thirty-six. They have a perfectly good rule, n times n, and their gaps run three, five, seven, nine, eleven. Having a rule is not the test. A constant gap is the test. Here is where the one-fewer-step idea earns its keep. A taxi charges a booking fee of two hundred, and then forty for every kilometre. After one kilometre you owe two hundred and forty. After two, two hundred and eighty. After three, three hundred and twenty.

7:25Those fares are an arithmetic progression: first term two hundred and forty, common difference forty. The fare after ten kilometres is six hundred. Now the trap. The rule can be written as two hundred plus forty n, and it is tempting to call two hundred the first term. It is not. The first term is two hundred and forty. Two hundred is what the meter reads before you have gone anywhere, and this sequence has no such position.

7:57Solve for where it would sit and you get position zero, which is not a place in a list. One more thing people get wrong, and it is about zero. Take twenty-one, eighteen, fifteen, and keep going down by three. Is minus eighty-one a term? Solve it: yes, it is the thirty-fifth. Is zero a term? Most people say no, because zero feels like it ought to be outside a progression rather than in one.

8:29But twenty-one, eighteen, fifteen, twelve, nine, six, three, zero. It is the eighth term. Nothing privileges zero. It either solves to a counting number or it does not, exactly like every other candidate. And the same solving answers the ordinary questions: for three, eight, thirteen, eighteen, the tenth term is forty-eight and the twenty-sixth is a hundred and twenty-eight. Finally, the reverse job. Somebody hands you two terms and wants the rest.

9:07A progression has twelve in position three, and a hundred and six in position fifty. The temptation is to divide the difference by fifty minus three. Do not. Ask instead how many additions stand between those two positions. Forty-seven. Those forty-seven additions carry you from twelve up to a hundred and six, a rise of ninety-four. So each one carries two. The common difference is two. Step back two additions from position three and the first term is eight. The twenty-ninth term is sixty-four.

9:45And the same counting settles a question in ordinary words. A salary starts at five hundred thousand and rises by twenty thousand every year. When does it reach seven hundred thousand? The gap is two hundred thousand, which is ten raises. Seven hundred thousand is the eleventh figure on the list, and it arrives after ten years, not eleven. Eleven terms, ten steps. It is the same fence post as the formula, wearing a different hat.