Arithmetic sequences

Count by 5s: 5, 10, 15, 20, \dots You have been doing this since primary school. A sequence is just an ordered list of numbers, and each number is called a term. What makes counting-by-5s special is that you add the same amount — 5 — to get from one term to the next. That single, simple idea is the whole of this lesson, dressed up in some new notation. We are not learning something alien; we are giving a precise name to a pattern you already feel.

To talk about terms precisely we label them by their position. The first term is t_1, the second is t_2, and the term in position n is t_n (read 't sub n'). So for 5, 10, 15, 20, \dots we have t_1 = 5, t_2 = 10, t_3 = 15. The little number is just an address telling you which term — it is not multiplication. Getting comfortable with this labelling is the key that unlocks every formula that follows.

A sequence is arithmetic when you get each term by adding a fixed number to the one before. That fixed number is the common difference, written d. You find it by subtracting any term from the one after it: d = t_2 - t_1 = t_3 - t_2, and so on. In 3, 7, 11, 15, \dots, d = 7 - 3 = 4. If d is positive the sequence climbs; if d is negative it falls (like 20, 17, 14, \dots with d = -3).

It is worth a moment to see what is not arithmetic. The list 2, 4, 8, 16, \dots looks patterned, but check the differences: 4-2 = 2, then 8-4 = 4, then 16-8 = 8 — they keep changing! Here you multiply by 2 each time, not add the same amount. That is a geometric sequence (next lesson). The test for arithmetic is always the same: subtract neighbouring terms and see if you get the same d every time.

Writing out t_1, t_2, t_3, \dots forever is hopeless if someone asks for the 100th term. So we build a shortcut. Starting from t_1, to reach t_2 you add d once; to reach t_3 you add d twice; to reach t_n you add d a total of (n-1) times. That gives the general term t_n = t_1 + (n-1)d. This one formula lets you jump straight to any term without listing them all.

This is the single most common slip, so let's nail it. You do not add d to get the first term — you are already there. You add d once to reach the second term, so by the time you reach the nth term you have added d exactly (n-1) times. Quick check: for 3, 7, 11, \dots, the 4th term should be 3 + (4-1)(4) = 3 + 12 = 15 ✓, which matches the list. Using n instead of n-1 would overshoot by one step every time.

Often you are told a couple of terms and must recover the whole sequence. Two facts you can lean on: the jump between any two terms is a whole number of d's, so if you know t_3 and t_7, then going from position 3 to position 7 is 7 - 3 = 4 steps, giving t_7 - t_3 = 4d. Solve that for d first, then substitute back into t_n = t_1 + (n-1)d to get t_1. Find d, then find t_1 — that order almost always works.

To ask 'is 100 a term of 4, 9, 14, \dots?' set the general term equal to 100 and solve for n: 4 + (n-1)5 = 100. The catch — and it is a real one — is that n must come out as a positive whole number, because there is no 'term number 20.2'. If n is a fraction or negative, the number simply is not in the sequence. Always check that your n is a counting number before saying 'yes'.

Here is a beautiful link to Grade 10. If you plot the points (position, term) — that is, (1, t_1), (2, t_2), (3, t_3), \dots — they line up perfectly straight! The common difference d is exactly the slope of that line: each step right (one position) rises by d. In fact t_n = t_1 + (n-1)d is just y = mx + b in disguise. Recognising this means every intuition you built about lines now works for arithmetic sequences too.