Sequences: nth terms, differences and recurrence rules

A sequence is an ordered list of numbers. Each number is a term, and its position matters.

For example, in

$3,7,11,15,\ldots$

the first term is $3$, the second is $7$, and the fourth is $15$. The dots mean that the sequence continues according to its rule.

Two descriptions of a sequence are especially important:

• a position-to-term rule gives a term directly from its position, such as $u_n=4n-1$;
• a term-to-term rule gives each new term from earlier terms, such as $u_{n+1}=u_n+4$.

Moving confidently between lists, formulae and verbal rules is essential preparation for A-level sequences, series and iteration.

Prerequisites

You should already be able to:

• substitute positive integers into algebraic expressions;
• simplify expressions and solve linear equations;
• expand and factorise quadratics;
• use negative numbers accurately.

Review function notation, algebraic manipulation and factorisation, linear equations or quadratic equations if needed.

Sequence notation

The symbol $u_n$ means the term in position $n$. Thus

$u_1=\text{first term},\qquad u_2=\text{second term},\qquad u_{10}=\text{tenth term}.$

The letter is only a name. Some questions use $a_n$, $x_n$ or another letter.

If

$u_n=5n+2,$

then substitute the required position for $n$:

$u_1=5(1)+2=7,\qquad u_4=5(4)+2=22.$

Misconception: $u_n$ is not $u\times n$

$u_n$ is a labelled term, not a product. Similarly, $u_{n+1}$ means the term after $u_n$. It does not mean $u_n+1$.

Generating terms from a position-to-term rule

A position-to-term rule, often called an nth-term rule, lets you find any term without generating all earlier terms.

Worked example 1: generate a sequence

The nth term is $u_n=3n-8$. Find the first five terms.

Use $n=1,2,3,4,5$:

$$\begin{aligned} u_1&=3(1)-8=-5,\\ u_2&=3(2)-8=-2,\\ u_3&=3(3)-8=1,\\ u_4&=3(4)-8=4,\\ u_5&=3(5)-8=7. \end{aligned}$$

Therefore the sequence is

$-5,-2,1,4,7,\ldots$

The constant difference is $3$, matching the coefficient of $n$.

Worked example 2: a non-linear rule

For $v_n=n^2-2n$, find $v_1$, $v_3$ and $v_{10}$.

$$\begin{aligned} v_1&=1^2-2(1)=-1,\\ v_3&=3^2-2(3)=3,\\ v_{10}&=10^2-2(10)=80. \end{aligned}$$

Use brackets when substituting a negative value, although sequence positions normally begin at $n=1$ and are positive integers.

Self-check 1

For $a_n=12-4n$:

1. write the first four terms;
2. find $a_{25}$;
3. state the common difference.

Answers

1. $8,4,0,-4$.
2. $a_{25}=12-4(25)=-88$.
3. The common difference is $-4$.

Term-to-term rules and recurrence relations

A recurrence relation defines terms using earlier terms. It must include enough starting information to begin the sequence.

For example,

$u_1=6,\qquad u_{n+1}=u_n+5$

means start at $6$ and repeatedly add $5$:

$6,11,16,21,\ldots$

Without $u_1=6$, the recurrence would describe many possible sequences, all increasing by $5$.

Worked example 3: generate terms recursively

Given

$x_1=20,\qquad x_{n+1}=\frac{x_n}{2}+3,$

find $x_2$, $x_3$ and $x_4$.

Apply the same rule to the latest term each time:

$$\begin{aligned} x_2&=\frac{x_1}{2}+3=\frac{20}{2}+3=13,\\ x_3&=\frac{x_2}{2}+3=\frac{13}{2}+3=\frac{19}{2},\\ x_4&=\frac{x_3}{2}+3=\frac{19}{4}+3=\frac{31}{4}. \end{aligned}$$

Keep exact fractions unless a decimal is requested.

Worked example 4: use two preceding terms

The Fibonacci-type recurrence

$f_1=2,\qquad f_2=5,\qquad f_{n+2}=f_{n+1}+f_n$

needs two starting terms because every new term uses two earlier terms.

$$\begin{aligned} f_3&=f_2+f_1=5+2=7,\\ f_4&=f_3+f_2=7+5=12,\\ f_5&=f_4+f_3=12+7=19. \end{aligned}$$

So the sequence begins $2,5,7,12,19,\ldots$.

Misconception: applying the rule to the position

In $x_{n+1}=2x_n-1$, double the previous term, not the position number. If $x_1=4$, then $x_2=2(4)-1=7$, not $2(1)-1$.

Self-check 2

Given $p_1=3$ and $p_{n+1}=2p_n-4$, find $p_2$, $p_3$, $p_4$ and $p_5$.

Answer

$p_2=2,qquad p_3=0,qquad p_4=-4,qquad p_5=-12.$

Linear sequences and constant first differences

A sequence is linear, or arithmetic, when the difference between consecutive terms is constant.

For

$7,12,17,22,\ldots$

the first differences are

$5,5,5,\ldots$

so its nth term has the form

$u_n=5n+c.$

The coefficient of $n$ is the common difference. To find $c$, compare with the sequence $5n$:

$n$	$1$	$2$	$3$	$4$
$5n$	$5$	$10$	$15$	$20$
required term	$7$	$12$	$17$	$22$

Every required term is $2$ larger, so

$\boxed{u_n=5n+2}.$

Equivalently, if the first term is $a$ and common difference is $d$, then

$\boxed{u_n=a+(n-1)d}.$

This works because reaching term $n$ from term $1$ requires $n-1$ equal steps, not $n$ steps.

Worked example 5: a decreasing sequence

Find the nth term of

$18,14,10,6,\ldots$

The common difference is $-4$, so begin with $-4n$:

$-4,-8,-12,-16,\ldots$

Add $22$ to obtain the required terms. Hence

$\boxed{u_n=-4n+22}.$

Check both the first term and the difference:

$u_1=-4+22=18,\qquad u_2-u_1=14-18=-4.$

Worked example 6: use $a+(n-1)d$

A linear sequence has first term $-3$ and common difference $7$. Find its nth term and its $50$th term.

$$\begin{aligned} u_n&=a+(n-1)d\\ &=-3+7(n-1)\\ &=7n-10. \end{aligned}$$

Therefore

$u_{50}=7(50)-10=340.$

Misconception: using the first term as the coefficient

In $4,10,16,22,\ldots$, the coefficient of $n$ is $6$, because the common difference is $6$. The formula is $6n-2$, not $4n+2$.

Self-check 3

Find the nth term of each linear sequence.

1. $9,13,17,21,\ldots$
2. $30,23,16,9,\ldots$
3. $-8,-3,2,7,\ldots$

Answers

1. Difference $4$: $u_n=4n+5$.
2. Difference $-7$: $u_n=-7n+37$.
3. Difference $5$: $u_n=5n-13$.

Is a number in a sequence?

To test whether a number is a term, set the nth-term formula equal to that number and solve for $n$. It is a term only if the solution is an allowed position, usually a positive integer.

Worked example 7: test membership of a linear sequence

The sequence has nth term $u_n=7n-4$. Is $150$ a term?

$$\begin{aligned} 7n-4&=150,\\ 7n&=154,\\ n&=22. \end{aligned}$$

Since $22$ is a positive integer, $150$ is the $22$nd term.

Is $100$ a term?

$7n-4=100\quad\Longrightarrow\quad n=\frac{104}{7}.$

This is not an integer, so $100$ is not a term.

Finding a numerical value of $n$ is not enough. A position such as $14.86$ has no meaning in an ordinarily indexed sequence.

Quadratic sequences and second differences

A quadratic sequence has an nth term of the form

$u_n=an^2+bn+c,$

where $a\ne0$. Its first differences are not constant, but its second differences are constant.

Consider

$2,7,14,23,34,\ldots$

Form a difference table:

$$\begin{array}{c|ccccc} n&1&2&3&4&5\\ \hline u_n&2&7&14&23&34\\ \text{first difference}&&5&7&9&11\\ \text{second difference}&&&2&2&2 \end{array}$$

For $u_n=an^2+bn+c$, the constant second difference is $2a$. Here $2a=2$, so $a=1$.

Why is the second difference $2a$?

The first difference is

$$\begin{aligned} u_{n+1}-u_n &=a(n+1)^2+b(n+1)+c-(an^2+bn+c)\\ &=2an+a+b. \end{aligned}$$

This is linear in $n$, so taking another difference gives $2a$. The constant $c$ disappears at the first difference, and the linear part contributes nothing to the second.

Finding a quadratic nth term

A reliable method is:

1. calculate first and second differences;
2. divide the second difference by $2$ to find $a$;
3. subtract the terms of $an^2$ from the original sequence;
4. find the linear nth term of the remainder.

Worked example 8: positive second difference

Find the nth term of

$4,11,22,37,56,\ldots$

The first differences are $7,11,15,19$, so the second difference is $4$. Therefore

$2a=4\quad\Longrightarrow\quad a=2.$

Subtract the sequence $2n^2$:

$n$	$1$	$2$	$3$	$4$	$5$
original	$4$	$11$	$22$	$37$	$56$
$2n^2$	$2$	$8$	$18$	$32$	$50$
remainder	$2$	$3$	$4$	$5$	$6$

The remainder has nth term $n+1$. Hence

$\boxed{u_n=2n^2+n+1}.$

Check a term not used mentally in the construction:

$u_4=2(4^2)+4+1=37.$

Worked example 9: negative second difference

Find the nth term of

$9,14,15,12,5,\ldots$

The first differences are $5,1,-3,-7$, and the second differences are all $-4$. Thus

$2a=-4\quad\Longrightarrow\quad a=-2.$

Subtract $-2n^2$, which is equivalent to adding $2n^2$:

$n$	$1$	$2$	$3$	$4$	$5$
original	$9$	$14$	$15$	$12$	$5$
$-2n^2$	$-2$	$-8$	$-18$	$-32$	$-50$
remainder	$11$	$22$	$33$	$44$	$55$

The remainder is $11n$, so

$\boxed{u_n=-2n^2+11n}.$

Misconception: the second difference is $a$

The second difference is $2a$, not $a$. If the second difference is $6$, the quadratic part is $3n^2$.

Self-check 4

Find the nth term of each quadratic sequence.

1. $3,8,15,24,35,\ldots$
2. $6,15,28,45,66,\ldots$
3. $10,13,14,13,10,\ldots$

Answers

1. First differences $5,7,9,11$ give second difference $2$, so subtract $n^2$. The remainder is $2,4,6,8,10$, giving $\boxed{u_n=n^2+2n}$.
2. Second difference $4$ gives $2n^2$. The remainder is $4,7,10,13,16$, giving $\boxed{u_n=2n^2+3n+1}$.
3. First differences $3,1,-1,-3$ give second difference $-2$, so subtract $-n^2$. The remainder is $11,17,23,29,35$, giving $\boxed{u_n=-n^2+6n+5}$.

Other patterns

Not every sequence is linear or quadratic.

Geometric patterns

In

$3,6,12,24,\ldots$

each term is multiplied by $2$. Its nth term is

$u_n=3\cdot2^{n-1}.$

The exponent is $n-1$ because the first term has undergone zero multiplications by $2$:

$u_1=3\cdot2^0=3.$

Constant ratios indicate a geometric pattern. Constant first differences indicate a linear pattern. Do not confuse the two.

Alternating patterns

The sequence

$5,-5,5,-5,\ldots$

can be written

$u_n=5(-1)^{n-1}.$

Powers of $-1$ alternate between $1$ and $-1$.

Pattern spotting is not proof

A finite list does not determine one unique continuation. For example, $1,2,3,\ldots$ might continue with $4$ under the rule $u_n=n$, but a different rule could match the first three terms and then change. In exam questions, use the stated rule or the intended standard pattern. When justifying a formula, verify it algebraically or against the defining recurrence.

Mixed self-check

1. Write the first four terms of $u_n=n^2-4n+6$.
2. Given $v_1=2$ and $v_{n+1}=3v_n+1$, find $v_4$.
3. Find the nth term of $41,35,29,23,\ldots$.
4. Is $250$ a term of the sequence with nth term $9n+7$?
5. Find the nth term of $5,12,23,38,57,\ldots$.
6. A sequence has first term $81$ and each term is one third of the preceding term. Write its first four terms and an nth-term rule.

Answers

1. $3,2,3,6$.
2. $v_2=7$, $v_3=22$, $v_4=67$.
3. The difference is $-6$, so $u_n=-6n+47$.
4. $9n+7=250$ gives $n=27$, so yes, it is the $27$th term.
5. First differences are $7,11,15,19$, so the second difference is $4$. Subtracting $2n^2$ leaves $3,4,5,6,7$, whose nth term is $n+2$. Therefore $u_n=2n^2+n+2$.
6. $81,27,9,3$ and $u_n=81\left(\dfrac13\right)^{n-1}$.

Checklist

Before moving on, check that you can:

• distinguish a position-to-term rule from a term-to-term rule;
• generate terms from an nth-term formula or recurrence relation;
• find the nth term of a linear sequence;
• recognise constant second differences and find a quadratic nth term;
• test whether a value occurs by solving for a positive integer position;
• distinguish additive, multiplicative and alternating patterns.

Next steps

Continue to sequences and recurrence relations for formal A-level notation and recursive definitions. Then study arithmetic sequences and series and geometric sequences and series, where terms are summed and infinite behaviour is considered.