Arithmetic sequences and series

An arithmetic sequence changes by the same amount from one term to the next. This fixed amount is the common difference $d$.

For example,

$7,11,15,19,\ldots$

is arithmetic because every term is $4$ more than the previous term, so $d=4$.

An arithmetic series is the result of adding terms of an arithmetic sequence:

$7+11+15+19+\cdots.$

The distinction matters. A sequence is a list; a series is a sum.

Prerequisites

You should be able to use nth terms and recurrence rules, solve linear equations, and rearrange formulae.

Recognising an arithmetic sequence

A sequence $(u_n)$ is arithmetic when

$u_{n+1}-u_n=d$

for every relevant positive integer $n$. The difference may be positive, negative or zero.

Sequence	Consecutive differences	Arithmetic?
$3,8,13,18,\ldots$	$5,5,5,\ldots$	Yes, $d=5$
$12,9,6,3,\ldots$	$-3,-3,-3,\ldots$	Yes, $d=-3$
$2,4,8,16,\ldots$	$2,4,8,\ldots$	No

Equal second differences do not make a sequence arithmetic. They indicate a quadratic sequence. Arithmetic sequences have equal first differences.

The nth term

Let the first term be $a$ and the common difference be $d$. Then

$$\begin{aligned} u_1&=a,\\ u_2&=a+d,\\ u_3&=a+2d,\\ u_4&=a+3d. \end{aligned}$$

Term $n$ is reached after $n-1$ steps from the first term, so

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

The factor is $n-1$, not $n$. Checking $n=1$ makes this unavoidable: the formula must give $u_1=a$.

Worked example 1: find a distant term

Find the $40$th term of $18,13,8,3,\ldots$.

Here $a=18$ and $d=-5$. Therefore

$$\begin{aligned} u_{40}&=a+(40-1)d\\ &=18+39(-5)\\ &=-177. \end{aligned}$$

The negative answer is reasonable because the sequence decreases by $5$ repeatedly.

Worked example 2: recover $a$ and $d$

In an arithmetic sequence, $u_6=19$ and $u_{15}=55$. Find $a$, $d$ and $u_{30}$.

Using $u_n=a+(n-1)d$ gives

$a+5d=19$

and

$a+14d=55.$

Subtract the first equation from the second:

$9d=36,$

so $d=4$. Substitute into $a+5d=19$:

$a+20=19,$

so $a=-1$. Hence

$u_{30}=-1+29(4)=115.$

Notice that the term numbers differ by $15-6=9$, matching the $9$ equal steps used to find $d$.

Worked example 3: is a number a term?

Determine whether $104$ is a term of the sequence $8,14,20,26,\ldots$.

Here $a=8$ and $d=6$. Set the nth term equal to $104$:

$8+6(n-1)=104.$

Then

$6(n-1)=96,qquad n-1=16,qquad n=17.$

Since $17$ is a positive integer, $104$ is the $17$th term.

If solving had produced, for example, $n=17.5$, the number would not be a term. Positions in a sequence are positive integers.

Self-check 1

An arithmetic sequence has $u_4=2$ and $u_{11}=30$.

1. Find $a$ and $d$.
2. Find $u_{50}$.
3. Is $101$ a term?

Answers

From $a+3d=2$ and $a+10d=30$, subtraction gives $7d=28$, so $d=4$ and $a=-10$.

$u_{50}=-10+49(4)=186.$

For $101$,

$-10+4(n-1)=101$

gives $n=28.75$. This is not an integer, so $101$ is not a term.

The sum of the first n terms

Write $S_n$ for the sum of the first $n$ terms. If the first term is $a$ and the last of these terms is $l$, write the sum forwards and backwards:

$$\begin{aligned} S_n&=a+(a+d)+(a+2d)+\cdots+l,\\ S_n&=l+(l-d)+(l-2d)+\cdots+a. \end{aligned}$$

Adding vertically makes every pair equal to $a+l$:

$2S_n=(a+l)+(a+l)+\cdots+(a+l).$

There are $n$ pairs, so

$\boxed{S_n=\frac n2(a+l)}.$

Since $l=u_n=a+(n-1)d$, an equivalent form is

$\boxed{S_n=\frac n2\bigl(2a+(n-1)d\bigr)}.$

These formulae say the same thing:

$\text{sum}=\text{number of terms}\times\text{mean of first and last terms}.$

Use $\frac n2(a+l)$ when the last term is known. Use $\frac n2(2a+(n-1)d)$ when $a$, $d$ and $n$ are known.

Worked example 4: sum a finite arithmetic series

Find the sum of the first $25$ terms of

$6+10+14+18+\cdots.$

Here $a=6$, $d=4$ and $n=25$:

$$\begin{aligned} S_{25}&=\frac{25}{2}\bigl(2(6)+(25-1)4\bigr)\\ &=\frac{25}{2}(12+96)\\ &=1350. \end{aligned}$$

As a check, the last term is $6+24(4)=102$, so the mean term is $(6+102)/2=54$. Then $25\times54=1350$.

Worked example 5: count the terms first

Find

$17+24+31+\cdots+346.$

The last term is given, but the number of terms is not. First solve

$17+(n-1)7=346.$

Thus

$7(n-1)=329,qquad n-1=47,qquad n=48.$

Now use the first and last terms:

$S_{48}=\frac{48}{2}(17+346)=24(363)=8712.$

Do not use $346/7$ to count the terms. The sequence does not begin at $7$ or $0$.

Worked example 6: sum between two positions

For $u_n=5n-2$, find the sum from $u_{12}$ to $u_{30}$ inclusive.

There are

$30-12+1=19$

terms. The extra $1$ is essential because both endpoints are included. Also,

$u_{12}=58,qquad u_{30}=148.$

These selected consecutive terms still form an arithmetic sequence, so

$\text{sum}=\frac{19}{2}(58+148)=1957.$

An alternative is $S_{30}-S_{11}$, not $S_{30}-S_{12}$.

Solving equations involving an arithmetic sum

Because $S_n$ contains both $n$ and $n-1$, an equation for the number of terms is usually quadratic. Solve it algebraically, then enforce the fact that $n$ is a positive integer.

Worked example 7: find how many terms were added

The sequence $4,7,10,13,\ldots$ has first term $4$ and difference $3$. Find $n$ if $S_n=175$.

$$\begin{aligned} 175&=\frac n2\bigl(2(4)+(n-1)3\bigr)\\ 350&=n(3n+5)\\ 0&=3n^2+5n-350\\ 0&=(3n+35)(n-10). \end{aligned}$$

Thus $n=10$ or $n=-35/3$. Reject the negative value because a number of terms must be a positive integer. Therefore

$\boxed{n=10}.$

Check: the tenth term is $31$, and $10(4+31)/2=175$.

Worked example 8: find an unknown common difference

An arithmetic sequence begins with $12$. The sum of its first $20$ terms is $810$. Find the common difference.

$$\begin{aligned} 810&=\frac{20}{2}\bigl(2(12)+19d\bigr)\\ 81&=24+19d\\ 57&=19d\\ d&=3. \end{aligned}$$

The last term is then $12+19(3)=69$, and $20(12+69)/2=810$ confirms the result.

Self-check 2

1. Find $S_{40}$ for the arithmetic sequence with $a=15$ and $d=-2$.
2. Find the sum $9+14+19+\cdots+204$.
3. The sequence $7,11,15,\ldots$ has $S_n=462$. Find $n$.

Answers

1. $S_{40}=\frac{40}{2}(30+39(-2))=20(-48)=-960$.
2. $9+(n-1)5=204$ gives $n=40$. Therefore $S_{40}=\frac{40}{2}(9+204)=4260$.
3. $462=\frac n2(14+4(n-1))$, so $2n^2+5n-462=0$. Factorising gives $(2n+33)(n-14)=0$. Hence $n=14$; reject $n=-33/2$.

Arithmetic sequences in context

The common difference represents a constant additive change. Check that this is a sensible model before using arithmetic formulae. Constant percentage change produces a geometric sequence instead.

Worked example 9: rows of seats

A theatre has $18$ seats in its first row. Each successive row has $3$ more seats than the row before. There are $24$ rows. Find the total number of seats.

The row sizes form an arithmetic sequence with $a=18$, $d=3$ and $n=24$:

$$\begin{aligned} S_{24}&=\frac{24}{2}\bigl(2(18)+23(3)\bigr)\\ &=12(105)\\ &=1260. \end{aligned}$$

Therefore the theatre has $1260$ seats. State the unit and answer the question in context.

Worked example 10: identify a model’s limit

A reservoir contains $5000$ litres and loses $120$ litres each day. Taking the initial amount as $u_1$, the model is

$u_n=5000-120(n-1).$

The model predicts a non-positive amount when

$5000-120(n-1)\leq0.$

This gives $n\geq42.\overline{6}$, so the first integer position with a non-positive prediction is $n=43$. In reality the reservoir cannot contain a negative volume, so the arithmetic model must stop once it empties. Formulae do not remove the need to interpret their domain.

Common misconceptions

• Sequence and series: $u_n$ is one term; $S_n$ is the sum of the first $n$ terms.
• Number of steps: from term $1$ to term $n$ there are $n-1$ differences.
• Negative difference: a decreasing arithmetic sequence has $d<0$. Keep the sign when substituting.
• Inclusive counting: terms from position $p$ to position $q$ inclusive number $q-p+1$.
• Invalid roots: a term number must be a positive integer. Reject negative, zero or non-integer solutions.
• Rounding: arithmetic sequence questions usually permit exact working. Do not round an exact common difference unnecessarily.
• Infinite sums: a non-constant arithmetic series does not have a finite sum to infinity. The terms do not approach zero.

Mixed self-check

An arithmetic sequence has fifth term $18$ and twelfth term $46$.

1. Find its first term and common difference.
2. Find the sum of its first $30$ terms.
3. Find the sum from its eighth term to its twentieth term inclusive.

Answers

The term conditions are

$a+4d=18,qquad a+11d=46.$

Subtracting gives $7d=28$, so $d=4$ and $a=2$.

$S_{30}=\frac{30}{2}\bigl(2(2)+29(4)\bigr)=15(120)=1800.$

Also $u_8=2+7(4)=30$ and $u_{20}=2+19(4)=78$. From position $8$ to position $20$ there are $20-8+1=13$ terms, so

$\text{required sum}=\frac{13}{2}(30+78)=702.$

Next steps

Learn how constant multiplication produces geometric sequences and series, write sums compactly using sigma notation, and apply both families in sequence and series models. For sequences defined from earlier terms, continue to sequences and recurrence relations.