1.2 Arithmetic sequences and series

A sequence is a list of numbers that follow a pattern. The simplest pattern is to add the same amount each time, as in the seats in a theatre where each row has two more seats than the row in front, or a salary that rises by a fixed amount every year. Sequences of this kind are called arithmetic sequences. They describe any quantity that changes by equal steps, and their sums answer questions such as how many seats the whole theatre holds or how much is earned over a whole career.

Sequences and their notation

A sequence is an ordered list of numbers. Each number in the list is a term, and its position matters: the sequence $2, 5, 8, 11$ is different from $11, 8, 5, 2$. A sequence can be finite, ending after a certain number of terms, or infinite, continuing for ever, which is shown by three dots: $2, 5, 8, 11, \ldots$

The subscript in $u_n$ gives the position of a term, not its value. In the sequence $2, 5, 8, 11, \ldots$, the first term is $u_1 = 2$ and the fourth term is $u_4 = 11$. The term after $u_n$ is $u_{n+1}$, and the term before it is $u_{n-1}$.

Two ways to define a sequence

A sequence can be given by a formula for the $n$th term, sometimes called the general term. Substituting $n = 1, 2, 3, \ldots$ produces the terms in order, and any term can be found directly. For example, $u_n = 3n - 1$ gives $u_1 = 2$, $u_2 = 5$, $u_3 = 8$ and $u_{100} = 299$.

A sequence can also be given by a recursive formula, which describes how to get each term from the one before, together with a first term. The recursive formula $u_{n+1} = u_n + 3$ with $u_1 = 2$ produces the same sequence $2, 5, 8, 11, \ldots$ A recursive rule shows clearly how the sequence grows, but finding $u_{100}$ from it means working through all the terms before it.

Arithmetic sequences

In the sequences $2, 5, 8, 11, \ldots$ and $16, 12, 8, 4, \ldots$ the gap between consecutive terms is always the same: $+3$ in the first and $-4$ in the second. Any sequence with this property is an arithmetic sequence, and the fixed gap is its common difference.

The common difference can be positive (an increasing sequence), negative (a decreasing sequence) or zero (a constant sequence such as $5, 5, 5, \ldots$). In recursive form, every arithmetic sequence satisfies $u_{n+1} = u_n + d$. To decide whether a list of numbers is arithmetic, subtract each term from the next one: the sequence is arithmetic only if every difference is the same.

The nth term of an arithmetic sequence

Writing out an arithmetic sequence in terms of $u_1$ and $d$ shows a pattern. To reach the second term, $d$ is added once; to reach the third term, it is added twice; and so on.

In every column, the number of $d$'s is one less than the position of the term, because the first term has no differences added to it. This gives the formula for the $n$th term.

Arithmetic sequences and straight lines

Expanding $u_n = u_1 + (n-1)d$ gives $u_n = dn + (u_1 - d)$, which has the form $y = mx + c$ with gradient $m = d$. Plotting the terms against their positions therefore gives points that lie on a straight line. The sequence is a linear function whose domain is the positive integers, so the graph is a set of separate points rather than a continuous line.

This link explains the shape of every arithmetic sequence: a positive $d$ gives points rising at a constant rate, a negative $d$ gives points falling at a constant rate, and $d = 0$ gives a horizontal row of points. Arithmetic sequences are the discrete version of the linear models in 2.1, and they represent change that happens by equal steps.

Finding the position of a term and the number of terms

The formula $u_n = u_1 + (n-1)d$ links four quantities. When three of them are known, the fourth can be found. Setting $u_n$ equal to a given value and solving for $n$ shows whether that value is in the sequence: it is a term only if $n$ is a positive integer.

The same idea counts the terms in a finite sequence. The last term is $u_n$, so solving $u_1 + (n-1)d = \text{last term}$ gives the number of terms $n$.

Finding the first term and the common difference

When two terms of an arithmetic sequence are given, each one gives an equation in $u_1$ and $d$. Solving the two equations simultaneously finds both unknowns, and then any other term or sum can be found.

Questions sometimes give three terms in algebraic form and ask for the value of an unknown. Because consecutive terms have equal differences, the middle term of any three consecutive terms is the mean of the other two: if $a, b, c$ are consecutive terms then $b - a = c - b$, so $2b = a + c$.

Arithmetic series

Adding the terms of a sequence gives a series. The sum of the first $n$ terms is written $S_n$, so for the sequence $2, 5, 8, 11, \ldots$ the sum $S_4 = 2 + 5 + 8 + 11 = 26$. When the sequence being added is arithmetic, the sum is an arithmetic series.

Deriving the sum formula

Adding many terms one at a time is slow, but arithmetic series have a shortcut. Consider $1 + 2 + 3 + \cdots + 100$. Pairing the first and last terms gives $1 + 100 = 101$, the second and second-last give $2 + 99 = 101$, and every such pair has the same total. There are 50 pairs, so the sum is $50 \times 101 = 5050$.

The same idea can be seen in a picture. Two copies of the staircase $1 + 2 + 3 + 4 + 5$, one turned upside down, fit together to make a rectangle.

The general proof writes the sum forwards and backwards. Each term in the forward sum is $d$ larger than the one before, and each term in the backward sum is $d$ smaller, so when the two lines are added, every column has the same total, $u_1 + u_n$.

There are $n$ columns, so $2S_n = n(u_1 + u_n)$. Dividing by 2 gives the first form of the sum formula. Substituting $u_n = u_1 + (n-1)d$ gives the second form.

The first form says that the sum is the number of terms multiplied by the mean of the first and last terms. It is the quicker form when the last term is known. The second form is used when the common difference is known instead.

Finding the number of terms from a sum

When the sum is known and $n$ is not, the sum formula becomes a quadratic equation in $n$. Solve it with the quadratic formula, by factorising or with a GDC, and keep only a positive integer solution. In questions that ask when a total first passes a value, solve the equation, then check the whole numbers on either side.

Finding a term from a formula for the sum

Sometimes the formula for $S_n$ is given and a term is needed. Since $S_n$ is the sum of the first $n$ terms and $S_{n-1}$ is the sum of the first $n - 1$, their difference is the $n$th term on its own:

Sigma notation

Long sums are written compactly with sigma notation. The symbol $\Sigma$ is the Greek capital letter sigma, for "sum". A variable, usually $k$ (sometimes $r$ or $i$), counts through whole numbers from a lower limit to an upper limit, and the expression after $\Sigma$ gives the term for each value of $k$.

For example, $\sum_{k=1}^{4} (3k - 1) = 2 + 5 + 8 + 11 = 26$. The number of terms in $\sum_{k=a}^{b}$ is $b - a + 1$, because both limits are included. So $\sum_{k=5}^{30}$ has $26$ terms, not $25$.

A GDC evaluates sigma notation directly, either with a summation template ($\Sigma$) or by summing a generated list (a "sum" of a "sequence" command). This is useful for checking a hand calculation, and for sums that do not start at $k = 1$.

Applications of arithmetic sequences

Many situations change by a fixed amount at each step: rows of seats that each hold a few more than the row in front, salaries with a fixed yearly rise, stacks in which each layer has one fewer item than the layer below. In each case the first step is to identify $u_1$, $d$ and what $n$ counts, and the second is to decide whether the question asks for a single term ($u_n$) or a total ($S_n$).

Salaries and simple interest

A salary scale with a fixed annual increment is an arithmetic sequence: $u_1$ is the starting salary, $d$ is the yearly rise and $u_n$ is the salary in year $n$. The total earned over the first $n$ years is $S_n$. In the same way, simple interest adds the same amount of interest each year, so the balance of a simple-interest account is an arithmetic sequence. Compound interest, where the balance is multiplied by a fixed factor, gives a geometric sequence instead (see 1.4).

Stacking problems

Objects stacked in layers, with one fewer in each layer, give arithmetic series. When the total number of objects is known, the number of layers comes from solving a quadratic equation in $n$.

Models that are only approximately arithmetic

Real data rarely follow a pattern exactly. When the differences between successive values are roughly equal, an arithmetic sequence can still be a useful model. A sensible estimate of the common difference is the mean of the differences, which equals $\frac{\text{last value} - \text{first value}}{\text{number of gaps}}$. The model can then be used to predict, but its predictions need interpreting with care.

The yearly increases in this (hypothetical) data set vary between $0.29$ m and $0.33$ m, so the heights are close to arithmetic but not exactly so. The mean increase is $\frac{2.75 - 1.20}{5} = 0.31$ m per year, which gives the model $h_n = 1.20 + 0.31(n - 1)$.

Using a model to estimate between known values (interpolation) is usually reliable when the model fits well. Using it beyond the data (extrapolation) is less reliable, and the further the prediction goes, the less trust it deserves. A complete answer to a modelling question states whether the prediction is interpolation or extrapolation and gives a reason based on the context.

Exam-style questions on arithmetic sequences

Arithmetic sequences appear in short questions on both papers and as the opening parts of longer questions. The questions below show typical structures: two given terms, a "show that" for $S_n$, a quadratic for $n$, and a comparison in context that needs a GDC.

Self-check

• To do:

  I can write down terms of a sequence from a formula for $u_n$ or from a recursive formula.

• To do:

  I can decide whether a sequence is arithmetic and find its common difference.

• To do:

  I can explain where $u_n = u_1 + (n-1)d$ comes from and use it to find a term, a position or the number of terms.

• To do:

  I can find $u_1$ and $d$ from two terms by solving simultaneous equations.

• To do:

  I can derive the sum formula by writing the series forwards and backwards.

• To do:

  I can find $S_n$, and find $n$ from a given sum, including "first exceeds" questions.

• To do:

  I can read, write and evaluate sums in sigma notation, by formula and with a GDC.

• To do:

  I can model a real situation with an arithmetic sequence and comment on whether a prediction is reliable.

For the 15th term, substitute $n = 15$.

(b) Start from $u_1 = 7$ and subtract $2.5$ each time.

(b) Substitute $u_1 = 7$ and $d = 4$ and simplify.

Check: $u_1 = 4 + 3 = 7$ and $u_{20} = 80 + 3 = 83$, as in (a).

Since $75$ is a positive integer, $301$ is the 75th term: $u_{75} = 301$. (b) Set the $n$th term equal to 150.

There is no 37.25th term, because $n$ must be a positive integer. So $150$ is not a term. It lies between $u_{37} = 149$ and $u_{38} = 153$.

Subtract equation (1) from equation (2) to eliminate $u_1$.

Substitute into (1): $u_1 + 16 = 17$, so $u_1 = 1$. Then

A shortcut for $d$: going from $u_5$ to $u_{12}$ adds $12 - 5 = 7$ differences, so $7d = 45 - 17 = 28$.

The terms are $2(3) + 1 = 7$, $5(3) - 2 = 13$ and $4(3) + 7 = 19$, so $d = 6$. The differences are equal, which confirms the answer.

(b) The series has $u_1 = 50$ and $d = -3$, and the last term is $-37$. First find how many terms there are.

Now use the form with the last term.

Subtract.

Check the first term separately: $u_1 = S_1 = 3 + 2 = 5$, and $6(1) - 1 = 5$, so the formula holds for $n = 1$ too. Then

which is constant, so the sequence is arithmetic with $d = 6$.

(b) The sequence $7, 11, 15, \ldots$ has $u_1 = 7$ and $d = 4$, so $u_k = 7 + 4(k-1) = 4k + 3$. The last term gives the upper limit.

Then $u_1 = 11 - 3 \times 4 = -1$. (b) Substitute into the formula from the booklet and simplify, showing each step.

(c) "Hence" means using the result from (b).

So $n = 25$ or $n = -23.5$. Since $n$ must be a positive integer, $n = 25$.