The nth Term of a Linear Sequence

The nth term of a linear sequence, including going down; pattern formulas like C = 3P − 2; whether a pattern is possible; the (n + 1)th term and fraction sequences.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 72 real past-paper questions below

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The nth Term Is a Rule for Position

The nth term turns a position n into the term. Put in n = 1 for the 1st term, n = 50 for the 50th.

n	1	2	3	4
Term	10	16	22	28

Rule

Here the nth term is $6 n + 4$ : the difference, 6, times n, then add 4.

Key idea

Linear means the gaps are all the same. For gaps that grow, see Quadratic and Cubic Sequences.

Questions also give you the nth term and ask for terms. Put each position in for n.

Worked example

1
$17 - 3 n$ — The question: write down the first three terms and the 10th term.
2
$17 - 3 \times 1 = 14$ — The 1st term: put n = 1. Multiply before you take away, so 3 × 1 = 3, then 17 − 3 = 14.
3
$17 - 3 \times 2 = 11$ — The 2nd term: put n = 2. 3 × 2 = 6, and 17 − 6 = 11.
4
$17 - 3 \times 3 = 8$ — The 3rd term: put n = 3. The terms go down by 3 each time, because of the −3 in front of n.
5
$17 - 3 \times 10 = - 13$ — The 10th term: put n = 10. 17 − 30 = −13. Terms can be negative.

Key idea

When the n part has a minus, the terms get smaller for ever. So the greatest term is the 1st one: here 14.

Finding It, Even When It Goes Down

“Find the nth term” is a 2-mark question. Find the difference, multiply it by n, then adjust. Here the sequence goes down, so watch the signs.

Worked example

1
$33, 28, 23, 18$ — The question: find the nth term. Each term is smaller than the one before.
2
$28 - 33 = - 5$ — Find the difference: take each term from the next one. Every gap is −5, so the sequence is linear.
3
$- 5 n$ — The difference goes in front of n. Keep its minus sign, because the sequence goes down.
4
$- 5 n : - 5, - 10, - 15, - 20$ — Put n = 1, 2, 3, 4 into −5n. Write these under the sequence to compare.
5
$33 - (- 5) = 38$ — Compare the first terms. Taking away a negative adds, so each term is 38 more than −5n.
6
$38 - 5 n$ — Put the two parts together. This is the same as −5n + 38.
7
$n = 2 : 38 - 5 \times 2 = 28$ — Check with n = 2. It gives 28, the 2nd term, so the rule is right.

Common mistake

Writing 5n + 28. A sequence that goes down must have −5n.

Key idea

One mark is for the −5n and one for the 38. If you are not sure of the number on the end, still write the −5n part.

Shortcut: The nth Term Formula

Textbooks also print a formula for the nth term. It gives the same answer, and it is quickest when you know the first term and the difference.

Formula

nth term = a + (n - 1) d

$a$ is the first term, $d$ is the difference (with its sign), and $n$ is the position of the term.

Worked example

1
$7, 11, 15, 19$ — The question: find the nth term.
2
$a = 7, d = 4$ — Read off the first term, 7, and the difference, 4: 11 − 7 = 4.
3
$7 + (n - 1) \times 4$ — Put a and d into the formula.
4
$7 + 4 n - 4$ — Multiply out the bracket: 4 × n = 4n and 4 × (−1) = −4.
5
$4 n + 3$ — Collect the numbers: 7 − 4 = 3. Same answer as the difference method.

When it goes down, d is negative. Here is the sequence from the last card.

Worked example

1
$a = 33, d = - 5$ — 33, 28, 23, 18: the first term is 33 and the difference is −5.
2
$33 + (n - 1) (- 5)$ — Put a and d in. Keep −5 in brackets so its minus is not lost.
3
$33 - 5 n + 5$ — Multiply out: (−5) × n = −5n, and (−5) × (−1) = +5, because minus times minus is plus.
4
$38 - 5 n$ — 33 + 5 = 38. The same answer as the last card.

Common mistake

Using n instead of n − 1. The 1st term has no steps added yet, so the nth term has only n − 1 steps of d.

Patterns and Formulas

Pattern questions use other letters. “Write a formula for C in terms of P” is the nth term with P instead of n.

Worked example

1
$3, 5, 7, 9$ — Counters in Patterns 1 to 4. Count them first.
2
$difference = 2$ — Each pattern has 2 more counters than the last, so the formula starts with 2P.
3
$2 P : 2, 4, 6, 8$ — Work out 2P for Patterns 1 to 4 and compare.
4
$3 - 2 = 1$ — Each count is 1 more than 2P.
5
$C = 2 P + 1$ — Write “C =” in front. The question asked for a formula for C.
6
$C = 2 \times 50 + 1 = 101$ — Pattern 50: put P = 50. 2 × 50 = 100, plus 1 is 101 counters.

Common mistake

Writing 2P + 1 without “C =”. A formula needs both sides.

Is There Enough?

Some questions give a limit: a number of counters you have, or “fewer than 150”.

Worked example

1
$C = 5 P - 2$ — The question: Mia has 95 counters. Can she make Pattern 20?
2
$C = 5 \times 20 - 2 = 98$ — Put P = 20. Pattern 20 needs 98 counters.
3
$98 > 95$ — 98 is more than the 95 she has, so the answer is no.

The largest pattern that uses fewer than 150 counters:

Worked example

1
$5 P - 2 < 150$ — Write the rule, then “less than 150”.
2
$5 P < 152$ — Add 2 to both sides.
3
$P < 30.4$ — Divide both sides by 5.
4
$P = 30$ — A pattern number must be a whole number, so take the whole number just below 30.4.
5
$5 \times 30 - 2 = 148$ — Check: Pattern 30 uses 148, under 150. Pattern 31 would use 153, too many.

Key idea

Give the reason with a number: “No, because Pattern 20 needs 98 counters.” Yes or no on its own scores nothing.

The (n + 1)th Term

The (n + 1)th term is the term just after the nth term. To find it, put (n + 1) where n was.

Worked example

1
$6 n + 1$ — The question: the nth term is 6n + 1. Find the (n + 1)th term.
2
$6 (n + 1) + 1$ — Replace n with (n + 1). Keep the bracket: the 6 multiplies all of it.
3
$6 n + 6 + 1$ — Multiply out the bracket: 6 × n = 6n and 6 × 1 = 6.
4
$6 n + 7$ — Collect the numbers: 6 + 1 = 7.
5
$n = 1 : 6 + 7 = 13$ — Check: with n = 1 it should give the 2nd term, and 6 × 2 + 1 = 13. It does.

Two terms next to each other. The kth and (k + 1)th terms of 6n + 1 add up to 212. Find k.

Worked example

1
$(6 k + 1) + (6 k + 7) = 212$ — Write the kth term, then the term after it, from the example above.
2
$12 k + 8 = 212$ — Collect like terms: 6k + 6k = 12k and 1 + 7 = 8.
3
$12 k = 204$ — Take 8 from both sides.
4
$k = 17$ — Divide both sides by 12. Check: 103 + 109 = 212.

Common mistake

Writing 6n + 1 + 1. Do not add 1 to the end: replace n with (n + 1).

Sequences of Fractions (Extended/Higher)

Some sequences are fractions. Do not look for one difference between the fractions: find a rule for the tops and a rule for the bottoms.

Worked example

1
$\frac{2}{5}, \frac{3}{8}, \frac{4}{11}, \frac{5}{14}$ — The question: find the nth term.
2
$2, 3, 4, 5$ — The tops go up by 1, and each is 1 more than n. So the top is n + 1.
3
$5, 8, 11, 14$ — The bottoms go up by 3, so start with 3n: 3, 6, 9, 12.
4
$5 - 3 = 2$ — Each bottom is 2 more than 3n, so the bottom is 3n + 2.
5
$\frac{n + 1}{3 n + 2}$ — Put the top rule over the bottom rule.
6
$n = 4 : \frac{5}{14}$ — Check with n = 4: 4 + 1 = 5 and 3 × 4 + 2 = 14. It matches the 4th term.

Is a fraction a term? Is 9/26 a term of this sequence?

Worked example

1
$\frac{n + 1}{3 n + 2} = \frac{9}{26}$ — Set the nth term equal to the fraction.
2
$26 (n + 1) = 9 (3 n + 2)$ — Cross-multiply: each top times the other bottom.
3
$26 n + 26 = 27 n + 18$ — Multiply out both brackets.
4
$8 = n$ — Take 26n and 18 from both sides.
5
$n = 8$ — n is a whole number, so yes: 9/26 is the 8th term. If n were not whole, it would not be a term.

Common mistake

Taking one fraction from the next. The gaps between fractions are not equal, so the difference method does not work on them.

Which Method? Read the Question

Find the difference first, then read what is asked.

You see	You do
“nth term”	Difference × n, then adjust
It goes down	The difference is negative
First term and difference	a + (n − 1)d
The nth term is given	Put n = 1, 2, 3 … in
“Formula for C in terms of P”	The nth term, written C = …
“Can she make Pattern 20?”	Work out Pattern 20, compare
“(n + 1)th term”	Replace n with (n + 1)
A sequence of fractions	A rule for the tops, one for the bottoms

Key idea

Check with n = 1 and n = 2 before moving on.

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