IB Analysis and Approaches HL · Arithmetic and geometric sequences · Section 1 of 3

Arithmetic sequences: general term, sum and sigma notation

3 worked examples · By Álvaro López

Key idea

An arithmetic sequence is a list of numbers in which you move from each term to the next by always adding the same amount. With two formulas, the one for the general term and the one for the sum, you can obtain any term or add hundreds of them without writing them out. It is the basis of this topic and appears in Paper 1 and in Paper 2, on its own or within context problems.

Sequences and the general term

A sequence is an ordered list of numbers: a first term, a second, a third, and so on. It is written $u_{1}, u_{2}, u_{3}, \dots, u_{n}, \dots$ , where the subscript $n \in Z^{+}$ gives the position.

There are two usual ways to describe a sequence:

With the general term, a formula that gives $u_{n}$ directly from $n$ . For example, $u_{n} = n^{2} + 1$ gives $2, 5, 10, 17, \dots$ and lets you calculate $u_{50} = 2501$ without going through the previous terms.
By recurrence, a rule that gives each term from the previous one, together with the first term. For example, $u_{1} = 3$ and $u_{n + 1} = 2 u_{n} - 1$ give $3, 5, 9, 17, \dots$ . To reach $u_{50}$ you must calculate the 49 previous terms.

A recursive definition always needs its starting point, because the rule alone does not say where to begin.

Sigma notation

To indicate a sum of terms, the Greek letter $Σ$ is used:

For example, $\sum_{r = 1}^{4} r^{2} = 1 + 4 + 9 + 16 = 30$ and $\sum_{r = 1}^{5} (2 r + 1) = 3 + 5 + 7 + 9 + 11 = 35$ . The lower limit need not be $1$ : $\sum_{r = 3}^{6} r = 3 + 4 + 5 + 6 = 18$ has four terms, namely $6 - 3 + 1$ .

Two rules follow from the sum itself: a constant can be taken outside, $\sum_{r = 1}^{n} c u_{r} = c \sum_{r = 1}^{n} u_{r}$ , and a sum of two expressions can be split into two sums. Also, adding a constant $n$ times gives $\sum_{r = 1}^{n} c = n c$ .

Arithmetic sequences

For example, $7, 4, 1, - 2, \dots$ is arithmetic with $u_{1} = 7$ and $d = - 3$ ; by contrast, $1, 2, 4, 8, \dots$ is not, because $2 - 1 \neq = 4 - 2$ . To check that a sequence is arithmetic you must see that all the consecutive differences coincide; two of them are not enough.

General term

Starting from $u_{1}$ and adding $d$ each time: $u_{2} = u_{1} + d$ , $u_{3} = u_{1} + 2 d$ , $u_{4} = u_{1} + 3 d$ . To reach $u_{n}$ there have been $n - 1$ steps of size $d$ .

Notice that the number of steps is $n - 1$ and not $n$ : the first term has not received any addition. If you write the formula expanded, $u_{n} = d n + (u_{1} - d)$ , you see that it is a first-degree expression in $n$ , with the difference $d$ as the coefficient of $n$ . That is why the points $(n, u_{n})$ lie on a straight line.

If two terms are known, $u_{p}$ and $u_{q}$ , the difference between them is $u_{q} - u_{p} = (q - p) d$ , so $d = \frac{u _{q} - u _{p}}{q - p}$ .

Sum of the first terms

When the terms of a sequence are added, a series is obtained; in an arithmetic sequence, an arithmetic series. The sum of the first $n$ terms is denoted $S_{n} = \sum_{r = 1}^{n} u_{r}$ .

There is a trick, attributed to the young Gauss, for adding $1 + 2 + \dots + 100$ : write the sum twice, the second time backwards, and add column by column.

$1100 + 2 + 99 + 3 + 98 + \dots + 99 + \dots + 2 + 100 + 1$

Each column adds to $101$ and there are $100$ columns, so twice the sum is $100 \cdot 101$ and the sum equals $5050$ . The same idea works for any arithmetic sequence, because moving one place along the first row adds $d$ and along the second row subtracts $d$ : all the columns add up to the same amount, $u_{1} + u_{n}$ .

Proof. We write $S_{n}$ twice:

$S_{n} S_{n} = u_{1} + (u_{1} + d) + \dots + (u_{n} - d) + u_{n} = u_{n} + (u_{n} - d) + \dots + (u_{1} + d) + u_{1}$

Adding column by column, each column gives $u_{1} + u_{n}$ and there are $n$ columns, so $2 S_{n} = n (u_{1} + u_{n})$ . The second form comes from substituting $u_{n} = u_{1} + (n - 1) d$ .

The first form has a simple reading: the sum is the number of terms times the average of the first and the last. Use it when you know $u_{n}$ ; use the second when you only know $u_{1}$ and $d$ .

Finding $u_{1}$ and $d$ from two pieces of data

An arithmetic sequence is determined by $u_{1}$ and $d$ . If the question gives two pieces of data (two terms, a term and a sum, the sum and the last term…), each piece is translated into an equation with $u_{1}$ and $d$ as unknowns, and the system is solved.

If the data are terms separated by several places, subtracting the two equations gives $(q - p) d = u_{q} - u_{p}$ directly.

How many terms to add to exceed a quantity

In many context problems you are asked how many terms are needed for the sum to exceed a value $K$ . Since $S_{n}$ is a second-degree expression in $n$ , the inequality $S_{n} > K$ is set up.

Worked examples

Example 1 Basic

Recognising a sequence and calculating terms

The sequence $5, 9, 13, 17, \dots$ is an arithmetic sequence.

a) Find the common difference $d$ .

b) Write the general term $u_{n}$ .

c) Calculate $u_{30}$ .

d) What position does the term $205$ occupy?

e) Write the recurrence rule that defines the sequence.

Solution

a) $d = 9 - 5 = 4$ . Checked with another pair: $13 - 9 = 4$ .

b) With $u_{1} = 5$ and $d = 4$ :

$u_{n} = 5 + (n - 1) \cdot 4 = 4 n + 1$

c) $u_{30} = 4 \cdot 30 + 1 = 121$ .

d) We look for $n$ with $u_{n} = 205$ : $4 n + 1 = 205$ , so $n = 51$ . The term $205$ occupies position $51$ . (By contrast, $100$ is not a term: $4 n + 1 = 100$ gives $n = \frac{99}{4}$ , which is not an integer.)

e) $u_{1} = 5$ and $u_{n + 1} = u_{n} + 4$ .

Example 2 Medium

Finding $u_{1}$ and $d$ from two terms, and summing

In an arithmetic sequence, $u_{4} = 17$ and $u_{9} = - 3$ .

a) Find the common difference $d$ .

b) Find the first term $u_{1}$ .

c) Write the general term $u_{n}$ .

d) Calculate the sum of the first twenty terms, $S_{20}$ .

Solution

a) Subtracting $u_{4}$ from $u_{9}$ , there are $9 - 4 = 5$ steps of size $d$ :

$u_{9} - u_{4} = 5 d ⟹ - 3 - 17 = 5 d ⟹ d = - 4$

b) As $u_{4} = u_{1} + 3 d$ : $17 = u_{1} + 3 (- 4)$ , so $u_{1} = 29$ . Check with $u_{9} = 29 + 8 (- 4) = - 3$ .

c) $u_{n} = 29 + (n - 1) (- 4) = 33 - 4 n$ .

d) First $u_{20} = 33 - 80 = - 47$ . Then

$S_{20} = \frac{20}{2} (u_{1} + u_{20}) = 10 (29 - 47) = - 180$

With the other form: $S_{20} = 10 (2 \cdot 29 + 19 \cdot (- 4)) = 10 (58 - 76) = - 180$ .

Example 3 Hard

Weekly savings: general term, sum and when a target is passed

A student is saving for a trip. In the first week she puts $12$ euros in a money box, and each following week she puts in $5$ euros more than the week before.

a) Find an expression for the amount $u_{n}$ deposited in week $n$ , and calculate the amount deposited in week $20$ .

b) Express the total saved after $n$ weeks as a sum in sigma notation, and find a formula for $S_{n}$ in terms of $n$ .

c) Find the least number of weeks needed to have more than $1000$ euros.

Solution

a) The deposits form an arithmetic sequence with $u_{1} = 12$ and $d = 5$ :

$u_{n} = 12 + (n - 1) \cdot 5 = 5 n + 7$

For week $20$ : $u_{20} = 5 \cdot 20 + 7 = 107$ euros.

b) The total after $n$ weeks is

$S_{n} = r = 1 \sum n (5 r + 7)$

Applying the formula with $u_{1} = 12$ and $d = 5$ :

$S_{n} = \frac{n}{2} (2 \cdot 12 + (n - 1) \cdot 5) = \frac{n}{2} (5 n + 19) = \frac{5 n ^{2} + 19 n}{2}$

Check with $n = 2$ : the actual sum is $12 + 17 = 29$ , and the formula gives $\frac{2 \cdot 29}{2} = 29$ .

c) We set up $S_{n} > 1000$ :

$\frac{5 n ^{2} + 19 n}{2} > 1000 ⟺ 5 n^{2} + 19 n - 2000 > 0$

The equation $5 n^{2} + 19 n - 2000 = 0$ has the positive root $n = \frac{- 19 + 40361}{10} \approx 18.19$ . The expression $S_{n}$ increases for $n > 0$ , so you need to go past that value, and as $n$ is an integer you round up: $n = 19$ .

Check: $S_{18} = \frac{5 \cdot 324 + 19 \cdot 18}{2} = 981 < 1000$ and $S_{19} = \frac{5 \cdot 361 + 19 \cdot 19}{2} = 1083 > 1000$ .

In the exam: marks are given for identifying $u_{1}$ and $d$ , for using the sum formula correctly and for setting up the inequality. The usual error in c) is rounding $18.19$ down to $18$ : after $18$ weeks there are $981$ euros, still less than $1000$ . In Paper 2 the root is obtained with the GDC, but it is worth writing the inequality and checking $S_{18}$ and $S_{19}$ .

Common mistakes

Summary

A sequence is described by its general term $u_{n}$ or by recurrence ( $u_{n + 1}$ in terms of $u_{n}$ , together with the first term). $\sum_{r = 1}^{n} u_{r} = u_{1} + \dots + u_{n}$ .
Arithmetic sequence: $u_{n + 1} = u_{n} + d$ , with a constant common difference $d$ .
General term: $u_{n} = u_{1} + (n - 1) d$ , and $d = \frac{u _{q} - u _{p}}{q - p}$ for any two terms.
Sum: $S_{n} = \frac{n}{2} (u_{1} + u_{n}) = \frac{n}{2} (2 u_{1} + (n - 1) d)$ , obtained by adding the series to itself written backwards.
With two pieces of data, set up a system for $u_{1}$ and $d$ . To find how many terms exceed a quantity, solve $S_{n} > K$ and round the result up.
The sum of the terms from $u_{p}$ to $u_{q}$ is $S_{q} - S_{p - 1}$ , with $q - p + 1$ terms.

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Arithmetic sequences: general term, sum and sigma notation

Sequences and the general term

Sigma notation

Arithmetic sequences

General term

Sum of the first terms

Finding u1​ and d from two pieces of data

How many terms to add to exceed a quantity

Worked examples

Common mistakes

Summary

Want to work on it together?

Finding $u_{1}$ and $d$ from two pieces of data