The nth term

A shortcut to any term

Suppose you want the $30$th term of $7, 11, 15, 19, \dots$ You could keep adding $4$ twenty-nine times. There is a faster way. Look at how each term is built.

The number of times $4$ is added is always one less than the term number. So the $n$th term is

This $a_n$ is also called the general term of the AP. For the $30$th term: $a_{30} = 7 + 29 \times 4 = 123$.

Four kinds of question

The formula links four quantities: $a$, $d$, $n$ and $a_n$. If you know three of them, you can find the fourth.

1. Find a term. The $18$th term of $-2, 3, 8, \dots$ has $a = -2$, $d = 5$: $a_{18} = -2 + 17 \times 5 = 83$.

2. Find which term has a value. Which term of $7, 11, 15, \dots$ is $211$?

1. Set up. $7 + (n - 1) \times 4 = 211$.
2. Undo the steps. $(n - 1) \times 4 = 204$, so $n - 1 = 51$.
3. Answer. $n = 52$: $211$ is the $52$nd term.

3. Test whether a number is a term. Is $300$ a term of the same AP? Then $(n - 1) \times 4 = 293$, and $n - 1 = 73.25$. A term number must be a whole number that is at least $1$, so $300$ is not a term.

4. Find $a$ or $d$. If the $11$th term of an AP with $d = 3$ is $35$, then $a + 10 \times 3 = 35$ and $a = 5$.

Counting from the end

A finite AP has a last term, called $l$. Reading it backwards gives another AP, with first term $l$ and common difference $-d$. So the $n$th term from the end is

A worked example. In the AP $3, 8, 13, \dots, 253$, find the $10$th term from the end.

1. Read off. $l = 253$ and $d = 5$.
2. Count back. $253 - 9 \times 5 = 253 - 45 = 208$.
3. Check. Counting back nine steps of $5$ from $253$ gives $208$.

Practise with new numbers

Try it in Python

This program finds the $30$th term of $7, 11, 15, \dots$ and then tests two numbers to see which term each one is.

Common mistakes

• Adding $d$ $n$ times. The $n$th term adds $d$ only $n - 1$ times: $a_{30} = a + 29d$.
• Forgetting the last $+1$. When you solve for $n$, you first find $n - 1$.
• Accepting a decimal term number. $n = 73.25$ means the number is not a term.
• Counting from the end with $+d$. Going backwards, you subtract $d$.

Remember these

• The $n$th term of an AP is $a_n = a + (n - 1)d$.
• Knowing three of $a$, $d$, $n$ and $a_n$ gives the fourth.
• A number is a term only if $n$ comes out a whole number of at least $1$.
• The $n$th term from the end is $l - (n - 1)d$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the $21$st term of $4, 10, 16, \dots$

2. Which term of $13, 16, 19, \dots$ is $70$?

3. Is $89$ a term of $2, 7, 12, \dots$?

4. The AP $1, 5, 9, \dots, 97$ has how many terms?

5. What is the $3$rd term from the end of $6, 11, 16, \dots, 81$?