The nth Term of a Linear Sequence

A taxi that charges £3 to get in and £2 a mile, seats added row by row across a theatre, savings that grow by the same amount every week — these all climb by a fixed step, and often you want a far-off value (the fare after 40 miles, the 100th row) without counting up to it one step at a time. That shortcut is exactly what the nth-term rule gives you.

A linear sequence goes up (or down) by the same amount each step — that constant gap is the common difference. A term-to-term rule says "add 2 to get the next one", but to reach, say, the 100th term you would have to add 2 again and again — ninety-nine times. The nth term rule fixes that: it is a single expression in the position n that you can substitute into to jump straight to any term, no matter how far along it is.

Think of n as a slot number: n = 1 is the 1st term, n = 2 is the 2nd, and n = 100 is the 100th. Feed the slot number into the rule and out pops the term that lives there.

Building the rule

Take the sequence 3, 5, 7, 9, \dots Each term is 2 more than the last, so the common difference is 2. That difference is the coefficient of n: the rule starts as 2n.

But 2n gives 2, 4, 6, 8, \dots — the 2 times table — which is 1 below every term we want. So we adjust the constant: add 1.

n\text{th term} = 2n + 1

Check it: the 4th term is 2 \times 4 + 1 = 9. And the 100th term is 2 \times 100 + 1 = 201 — no counting required.

The recipe

Every linear nth term has exactly the same shape — the common difference times n, then a fixed number nudged on top:

n\text{th term} = (\text{common difference}) \times n + (\text{adjustment})

So finding it is always two steps:

The gap is the coefficient of n. Find the common difference (how much you add each step) and write it in front of n. This builds a "times table" like 2n or 5n.
Fix the constant. Compare your times table with the real sequence. They differ by the same fixed amount everywhere — add (or subtract) that amount to land exactly on the sequence.

A neat shortcut for the adjustment: it is the "zeroth term" — the value you would get one step before the first term. For 3, 5, 7, 9, stepping back from 3 by 2 gives 1, which is exactly the +1.

Jumping to the 100th term

This is where the rule earns its keep. Once you have it, the 100th term is a single substitution — n = 100:

2(100) + 1 = 200 + 1 = 201

Want the 1000th? 2(1000) + 1 = 2001. The term-to-term rule would have made you add 2 nine hundred and ninety-nine times; the nth term rule does it in one line.

Worked examples

Example 1 — 4, 7, 10, 13, \dots

Common difference is 3, so start with 3n. That gives 3, 6, 9, 12 — each 1 short. Add 1:

n\text{th term} = 3n + 1

Check: n = 1 gives 3(1) + 1 = 4. The 100th term is 3(100) + 1 = 301.

Example 2 — a decreasing sequence 11, 9, 7, 5, \dots

Now the terms go down by 2, so the common difference is -2 and we start with -2n. That gives -2, -4, -6, -8 — every one is 13 too low. Add 13:

n\text{th term} = -2n + 13

Check: n = 1 gives -2(1) + 13 = 11. A falling sequence simply has a negative coefficient of n.

Example 3 — 5, 10, 15, 20, \dots

Common difference 5 gives 5n, which is already 5, 10, 15, 20 — bang on. The adjustment is 0, so the rule is just:

n\text{th term} = 5n

2n + 1 does not mean "add the position to twice something". The safe test is to substitute: n = 1 gives 2(1) + 1 = 3, which is indeed the first term. Always check by plugging in n = 1.
The common difference is the multiplier of n — not the first term. In 3, 5, 7 the rule is 2n + 1, not 3n: the 3 is where it starts, the 2 is how fast it grows.
The constant on the end is the zeroth term, not the first. It is what you would have one step before the sequence begins.

Suppose you lay tiles to build a path: the 1st picture uses 4 tiles, and every picture after that adds 3 more. The "add 3 each time" is the common difference, so it becomes the coefficient of n; the few extra tiles you started with become the constant. The path's rule is 3n + 1 — so picture number 50 would need 3(50) + 1 = 151 tiles, and you never had to build the first forty-nine.

can can can can can

A shopkeeper builds a display that starts with 2 cans on the floor and adds a shelf of 5 cans for each level up. Level n holds 5n + 2 cans — the 5 is the steady "per shelf" growth (the common difference) and the 2 is the unchanging base. The same recipe, whether it is tiles, cans, or numbers on a page.

See it built

Watch the rule come out of the table: line up each position n against its term, read the common difference as the coefficient, then nudge the constant until the rows match. Step through it.

See it grow