Number Patterns and the nth Term

Sequences, their terms and the common difference; finding an algebraic expression for the nth term of a number pattern or a pattern of figures; using it to find any term and to decide whether a number is a term.

Before you start

At primary school you continued number patterns such as 2, 5, 8, 11, … by finding the rule ("add 3 each time"), and patterns of figures made of sticks, dots or tiles by drawing the next figure and counting.

A letter stands for a number. $4n - 1$ means 4 times $n$ , then subtract 1. To evaluate it, substitute a number for the letter and work in the order of operations: when $n = 5$ , $4n - 1 = 4 \times 5 - 1 = 20 - 1 = 19$ .

Negative numbers can appear: when $n = 8$ , $23 - 3n = 23 - 3 \times 8 = 23 - 24 = -1$ .

Sequences and Their Terms

A number pattern such as 3, 7, 11, 15, … is called a sequence. The numbers are written in a definite order and each one is made from a rule. The dots "…" mean that the sequence goes on.

In the sequence 3, 7, 11, 15, … the first term is 3, the second term is 7, the third term is 11 and the fourth term is 15. We write $T_{1} = 3$ , $T_{2} = 7$ , $T_{3} = 11$ and $T_{4} = 15$ . The small number tells the position of the term: $T_{10}$ is the 10th term and $T_{50}$ is the 50th term.

The term in position $n$ , where $n$ stands for any of the whole numbers 1, 2, 3, …, is the nth term, written $T_{n}$ . Keep the two numbers apart: in $T_{2} = 7$ , the 2 is the position (the term number) and 7 is the term. $T_{2} = 7$ says that the second term is 7; it does not say that there are 7 terms.

Definitions

A sequence is a list of numbers in a definite order that follows a rule. Each number in a sequence is a term.

$T_{1}$ is the first term, $T_{2}$ the second term, and $T_{n}$ the nth term, where the term number $n$ is a positive whole number: 1, 2, 3, ….

To continue a sequence, find its rule by comparing each term with the term before it, check that the rule works for every term given, and then use it.

Example 1: Continuing a sequence

Write down the next two terms of each sequence, and say what the rule is.
(a) 5, 9, 13, 17, …
(b) 50, 44, 38, 32, …
(c) 3, 6, 12, 24, …
(d) 1, 4, 9, 16, …

Solution

(a) $9 - 5 = 4$ , $13 - 9 = 4$ and $17 - 13 = 4$ . Each term is 4 more than the term before it. The next terms are $17 + 4 = 21$ and $21 + 4 = 25$ .

(b) $44 - 50 = -6$ , $38 - 44 = -6$ and $32 - 38 = -6$ . Each term is 6 less than the term before it. The next terms are $32 - 6 = 26$ and $26 - 6 = 20$ .

(c) The differences are 3, 6 and 12, which are not the same. But $6 = 3 \times 2$ , $12 = 6 \times 2$ and $24 = 12 \times 2$ : each term is 2 times the term before it. The next terms are $24 \times 2 = 48$ and $48 \times 2 = 96$ .

(d) The differences are 3, 5 and 7: they go up by 2 each time. The next differences are 9 and 11, so the next terms are $16 + 9 = 25$ and $25 + 11 = 36$ . (These are $1 \times 1$ , $2 \times 2$ , $3 \times 3$ , …, the square numbers, which come again at the end of this lesson.)

(a) 21, 25 (add 4) (b) 26, 20 (subtract 6) (c) 48, 96 (multiply by 2) (d) 25, 36 (the differences go up by 2)

The Common Difference

In Example 1 (a) the same number, 4, is added each time. In (b) the same number is taken away each time: subtracting 6 is the same as adding $-6$ . Sequences like these are the ones whose nth term this lesson finds by reasoning.

Common difference

When each term of a sequence is found by adding the same number to the term before it, that number is the common difference of the sequence.

Common difference = any term − the term before it.

If the terms go up, the common difference is positive. If the terms go down, the common difference is negative.

Example 2: Finding the common difference

Which of these sequences have a common difference? State it where there is one.
(a) 4, 11, 18, 25, …
(b) 30, 26, 22, 18, …
(c) 2, 5, 9, 14, …
(d) 1.5, 2, 2.5, 3, …

Solution

(a) $11 - 4 = 7$ , $18 - 11 = 7$ , $25 - 18 = 7$ . The difference is the same each time: 7.

(b) $26 - 30 = -4$ , $22 - 26 = -4$ , $18 - 22 = -4$ . The difference is the same each time: $-4$ . The terms go down, so it is negative.

(d) $2 - 1.5 = 0.5$ , $2.5 - 2 = 0.5$ , $3 - 2.5 = 0.5$ . The difference is the same each time: 0.5.

(a) common difference 7 (b) common difference $-4$ (c) no common difference (d) common difference 0.5

Finding the nth Term

To find the 50th term of 3, 7, 11, 15, … we could keep adding 4 until we reach it, but that is slow and easy to get wrong. It is better to find an expression in $n$ for the nth term: a rule that turns any term number straight into its term.

The terms go up by 4 each time. So does $4n$ , the 4 times table. Put the two side by side:

Term number $n$	1	2	3	4
$4n$	4	8	12	16
Term $T_{n}$	3	7	11	15

Every term is 1 less than $4n$ . So the nth term is $T_{n} = 4n - 1$ .

Check it on two terms: $T_{2} = 4 \times 2 - 1 = 7$ and $T_{4} = 4 \times 4 - 1 = 15$ . Both agree with the sequence.

The nth term of a sequence with a common difference

1. Find the common difference, $d$ .

2. Write down $dn$ (the $d$ times table) for $n = 1, 2, 3, …$ under the terms.

3. Find the number that must be added to $dn$ to give each term. It is the same number for every term, and it is $T_{1} - d$ . Then $T_{n} = dn$ + that number (or $dn$ − a number, if each term is less than $dn$ ).

4. Check your expression on two terms of the sequence.

Why the number added is the same for every term

When the term number $n$ goes up by 1, the term changes by $d$ , and so does $dn$ . The two start a certain distance apart and change together, so they stay that same distance apart for every term. This is why the coefficient of $n$ in the nth term is always the common difference.

Example 3: A sequence that goes up

Find an expression for the nth term of the sequence 5, 8, 11, 14, …

Solution

The common difference is $8 - 5 = 3$ (and $11 - 8 = 3$ , $14 - 11 = 3$ ), so compare the terms with $3n$ .

For $n = 1, 2, 3, 4$ : $3n$ is 3, 6, 9, 12, and the terms are 5, 8, 11, 14. Each term is 2 more than $3n$ .

So $T_{n} = 3n + 2$ .

Check: $T_{3} = 3 \times 3 + 2 = 11$ and $T_{4} = 3 \times 4 + 2 = 14$ . Both are correct.

The nth term is $3n + 2$ .

Example 4: When each term is less than the times table

Find an expression for the nth term of the sequence 2, 8, 14, 20, …

Solution

The common difference is $8 - 2 = 6$ , so compare the terms with $6n$ .

For $n = 1, 2, 3, 4$ : $6n$ is 6, 12, 18, 24, and the terms are 2, 8, 14, 20. Each term is 4 less than $6n$ . (In the words of the rule: the number added is $T_{1} - d = 2 - 6 = -4$ .)

So $T_{n} = 6n - 4$ .

Check: $T_{2} = 6 \times 2 - 4 = 8$ and $T_{4} = 6 \times 4 - 4 = 20$ . Both are correct.

The nth term is $6n - 4$ .

The same reasoning works when the terms go down. The common difference is negative, so compare the terms with a negative multiple of $n$ .

Example 5: A sequence that goes down

Find an expression for the nth term of the sequence 20, 17, 14, 11, …

Solution

The common difference is $17 - 20 = -3$ (and $14 - 17 = -3$ , $11 - 14 = -3$ ), so compare the terms with $-3n$ .

For $n = 1, 2, 3, 4$ : $-3n$ is $-3$ , $-6$ , $-9$ , $-12$ , and the terms are 20, 17, 14, 11.

The first term is $20 - (-3) = 20 + 3 = 23$ more than $-3n$ . So is the second: $17 - (-6) = 17 + 6 = 23$ . Each term is 23 more than $-3n$ .

So $T_{n} = -3n + 23$ , which is written $T_{n} = 23 - 3n$ .

Check: $T_{2} = 23 - 3 \times 2 = 23 - 6 = 17$ and $T_{4} = 23 - 3 \times 4 = 23 - 12 = 11$ . Both are correct.

The nth term is $23 - 3n$ .

Finding a Term, and Deciding Whether a Number Is a Term

Once the nth term is known, any term can be found at once: substitute its term number for $n$ .

Example 6: Finding a given term

(a) Find the 50th term and the 100th term of 3, 7, 11, 15, …
(b) Find the 30th term of 20, 17, 14, 11, …

Solution

(a) The nth term is $4n - 1$ (found above).

$T_{50} = 4 \times 50 - 1 = 200 - 1 = 199$

$T_{100} = 4 \times 100 - 1 = 400 - 1 = 399$

(b) The nth term is $23 - 3n$ (Example 5).

$T_{30} = 23 - 3 \times 30 = 23 - 90 = -67$

(a) The 50th term is 199 and the 100th term is 399.
(b) The 30th term is $-67$ .

To decide whether a number is a term, read what the nth term says about every term. In 3, 7, 11, 15, … the nth term is $4n - 1$ , and $4n$ is a multiple of 4. So every term is 1 less than a multiple of 4, and a number is a term exactly when adding 1 to it gives one of the multiples of 4: 4, 8, 12, … (since $n$ is 1, 2, 3, …).

Example 7: Is it a term?

Is (a) 100, (b) 99 a term of the sequence 3, 7, 11, 15, …? If it is, which term is it?

Solution

Every term is $4n - 1$ : 1 less than 4 times its term number.

(a) $100 + 1 = 101$ . $101 \div 4 = 25$ remainder 1, so 101 is not a multiple of 4, and 100 is not 1 less than a multiple of 4.

Another way to see it: $T_{25} = 4 \times 25 - 1 = 99$ and $T_{26} = 4 \times 26 - 1 = 103$ . The sequence jumps from 99 to 103 and misses 100.

(b) $99 + 1 = 100 = 4 \times 25$ , a multiple of 4. So 99 is the term with $n = 25$ .

Check: $T_{25} = 99$ , found in (a).

(a) 100 is not a term of the sequence.
(b) 99 is a term: it is the 25th term.

Example 8: Is it a term? A sequence that goes down

The nth term of 20, 17, 14, 11, … is $23 - 3n$ . Is (a) $-40$ , (b) 0, (c) 26 a term of the sequence? If it is, which term is it?

Solution

Every term is 23 minus $3n$ . So 23 minus a term is $3n$ , which must be 3 times a term number: one of 3, 6, 9, ….

(a) $23 - (-40) = 23 + 40 = 63$ , and $63 = 3 \times 21$ . So $-40$ is the term with $n = 21$ . Check: $T_{21} = 23 - 3 \times 21 = 23 - 63 = -40$ .

(b) $23 - 0 = 23$ , and 23 is not a multiple of 3 ( $23 \div 3 = 7$ remainder 2). So 0 is not a term. Indeed $T_{7} = 23 - 21 = 2$ and $T_{8} = 23 - 24 = -1$ : the sequence jumps from 2 to $-1$ .

(c) $23 - 26 = -3 = 3 \times (-1)$ . The term number would have to be $-1$ , and term numbers are 1, 2, 3, …. So 26 is not a term. This is also clear from the sequence: it starts at 20 and only goes down.

(a) $-40$ is the 21st term.
(b) 0 is not a term.
(c) 26 is not a term.

Patterns of Figures

In a pattern of figures, the figure number is the term number $n$ , and the number of matchsticks (or dots, or tiles) in each figure is the term. Count them, put them in a table, and find the nth term as before. Then look at the drawing to see why the expression is right: each part of it should match a part of the figure.

Example 9: Squares made of matchsticks

Mei Ling makes a row of squares with matchsticks. Figure $n$ has $n$ squares.
(a) Find the number of matchsticks in Figure 4.
(b) Find, in terms of $n$ , the number of matchsticks in Figure $n$ , and explain your answer from the drawing.
(c) How many matchsticks are there in Figure 50?
(d) Is there a figure that uses exactly 100 matchsticks? If there is, which figure is it?

Solution

(a) Counting: Figure 1 has 4 matchsticks, Figure 2 has 7 and Figure 3 has 10. Figure 4 has one more square than Figure 3, which needs 3 more matchsticks: $10 + 3 = 13$ .

(b) The common difference is 3, so compare with $3n$ .

Figure number $n$	1	2	3	4
$3n$	3	6	9	12
Number of matchsticks	4	7	10	13

Each number of matchsticks is 1 more than $3n$ , so Figure $n$ has $3n + 1$ matchsticks. Check: Figure 2 has $3 \times 2 + 1 = 7$ and Figure 3 has $3 \times 3 + 1 = 10$ .

From the drawing: there is one matchstick at the left end. Each square then needs 3 more matchsticks (its top, its bottom and its right side). Figure $n$ has $n$ squares, so it has $1 + 3n = 3n + 1$ matchsticks.

(d) Take away the 1 matchstick at the left end: $100 - 1 = 99$ . The rest come in threes, one three for each square, and $99 = 3 \times 33$ . So Figure 33 uses exactly 100 matchsticks. Check: $3 \times 33 + 1 = 100$ .

(a) Figure 4 has 13 matchsticks.
(b) Figure $n$ has $3n + 1$ matchsticks.
(c) Figure 50 has 151 matchsticks.
(d) Yes: Figure 33 uses exactly 100 matchsticks.

Example 10: Tiles on a void deck

Mr Tan lays a path of blue tiles on a void deck and puts grey tiles all round it. Figure $n$ has $n$ blue tiles in a row.
(a) Find, in terms of $n$ , the number of grey tiles in Figure $n$ , and explain your answer from the drawing.
(b) How many grey tiles are needed for a path of 40 blue tiles?
(c) Mr Tan has exactly 50 grey tiles. How many blue tiles must the path have for him to use all of them?

Solution

(a) Counting the grey tiles: Figure 1 has 8, Figure 2 has 10, Figure 3 has 12. The common difference is 2, so compare with $2n$ : 2, 4, 6. Each number of grey tiles is 6 more than $2n$ , so Figure $n$ has $2n + 6$ grey tiles.

From the drawing: above each blue tile there is one grey tile and below it there is another, which makes $2n$ grey tiles. At each end of the path there is a column of 3 grey tiles, which makes 6 more. So there are $2n + 6$ grey tiles.

Check: Figure 3 has $2 \times 3 + 6 = 12$ grey tiles, as counted.

(b) $2 \times 40 + 6 = 80 + 6 = 86$

(c) The two ends always take 6 grey tiles, which leaves $50 - 6 = 44$ . These go 2 to each blue tile, one above it and one below, and $44 = 2 \times 22$ . So the path has 22 blue tiles. Check: $2 \times 22 + 6 = 50$ .

(a) Figure $n$ has $2n + 6$ grey tiles.
(b) 86 grey tiles are needed.
(c) The path must have 22 blue tiles.

Square, Cube and Triangular Numbers

Not every pattern has a common difference. Three patterns without one are worth knowing by name, with their nth terms.

Square numbers 1, 4, 9, 16, 25, …: Figure $n$ is a square of $n$ rows with $n$ dots in each row, so it has $n \times n = n^{2}$ dots.

Cube numbers 1, 8, 27, 64, 125, …: a cube built of small cubes, $n$ along each edge, is made of $n \times n \times n = n^{3}$ small cubes.

Triangular numbers 1, 3, 6, 10, 15, …: Figure $n$ is a triangle of dots with rows of 1, 2, 3, …, $n$ dots, so it has $1 + 2 + 3 + … + n$ dots.

Two copies of the same triangle, one of them turned upside down, fit together into $n$ rows with $n + 1$ dots in each row: row 1 of one triangle (1 dot) goes with row $n$ of the other ( $n$ dots), row 2 with row $n - 1$ , and so on. So the two triangles have $n(n + 1)$ dots together, and one triangle has half of $n(n + 1)$ dots: the nth triangular number is $\frac{n(n + 1)}{2}$ . For Figure 4: $\frac{4 \times 5}{2} = 10$ , and $1 + 2 + 3 + 4 = 10$ .

Pattern	First five terms	nth term $T_{n}$
Square numbers	1, 4, 9, 16, 25	$n^{2}$
Cube numbers	1, 8, 27, 64, 125	$n^{3}$
Triangular numbers	1, 3, 6, 10, 15	$\frac{n(n + 1)}{2}$

Other sequences are often made from these. Compare a sequence with the square numbers (or the cube numbers) the way you compared a sequence with $dn$ .

Example 11: Using the special patterns

(a) Find the 12th square number and the 6th cube number.
(b) Find the 10th triangular number.
(c) Find an expression for the nth term of the sequence 2, 5, 10, 17, 26, …

Solution

(a) $12^{2} = 12 \times 12 = 144$ and $6^{3} = 6 \times 6 \times 6 = 216$ .

(b) $\frac{10 \times 11}{2} = \frac{110}{2} = 55$ . Check by adding: $1 + 2 + 3 + … + 10 = 55$ .

(c) The differences are 3, 5, 7, 9, which are not the same, so there is no common difference. Compare with the square numbers 1, 4, 9, 16, 25: each term is 1 more. So $T_{n} = n^{2} + 1$ .

Check: $T_{3} = 3^{2} + 1 = 10$ and $T_{5} = 5^{2} + 1 = 26$ . Both are correct.

(a) The 12th square number is 144 and the 6th cube number is 216.
(b) The 10th triangular number is 55.
(c) The nth term is $n^{2} + 1$ .

Common Mistakes

Adding the common difference to $n$ . For 3, 7, 11, 15, … the nth term is not $n + 4$ (that gives $T_{1} = 5$ ). The common difference multiplies $n$ : the nth term is $4n - 1$ .
Using the first term as the number added. For 3, 7, 11, 15, … the nth term is not $4n + 3$ (that gives $T_{1} = 7$ , the second term). The number added is what turns $4n$ into the term: $T_{1} - 4 = 3 - 4 = -1$ .
Losing the minus sign when the terms go down. For 20, 17, 14, 11, … the common difference is $-3$ , so the nth term is $23 - 3n$ , not $3n + 17$ or $3n + 23$ . A sequence that goes down has a negative coefficient of $n$ .
Mixing up the term number and the term. "Is 100 a term of 3, 7, 11, …?" is not answered by working out $T_{100} = 399$ . Ask instead whether 100 can be $4n - 1$ for a positive whole number $n$ (1, 2, 3, …).
Checking only one term. For 3, 7, 11, … the expression $n + 2$ gives $T_{1} = 3$ , which is right, but $T_{2} = 4$ , which is wrong. Check on two terms.
Expecting every sequence to have a common difference. 1, 4, 9, 16, … has none; its nth term is $n^{2}$ . Work out the differences before you look for $dn$ .
Accepting a term number that is not 1, 2, 3, …. If the reasoning leads to a term number of 0, a negative number or a fraction, the number is not a term.

Practice

Work each question on paper before you open its solution. Check every nth term you find on two terms of the sequence.

Question 1

Write down the next two terms of each sequence and say what the rule is. Which of the sequences have a common difference?
(a) 7, 12, 17, 22, …
(b) 64, 32, 16, 8, …
(c) 45, 38, 31, 24, …

ShowHide solution

(a) $12 - 7 = 5$ , $17 - 12 = 5$ , $22 - 17 = 5$ : add 5 each time. Next terms: $22 + 5 = 27$ and $27 + 5 = 32$ .

(b) The differences are $-32$ , $-16$ , $-8$ , which are not the same. Each term is half of the term before it: divide by 2. Next terms: $8 \div 2 = 4$ and $4 \div 2 = 2$ .

(c) $38 - 45 = -7$ , $31 - 38 = -7$ , $24 - 31 = -7$ : subtract 7 each time. Next terms: $24 - 7 = 17$ and $17 - 7 = 10$ .

(a) 27, 32 (common difference 5)
(b) 4, 2 (divide by 2; no common difference)
(c) 17, 10 (common difference $-7$ )

Question 2

The nth term of a sequence is $5n - 2$ . Write down its first four terms, and find its 20th term.

ShowHide solution

$T_{1} = 5 \times 1 - 2 = 3$ , $T_{2} = 5 \times 2 - 2 = 8$ , $T_{3} = 5 \times 3 - 2 = 13$ , $T_{4} = 5 \times 4 - 2 = 18$ .

$T_{20} = 5 \times 20 - 2 = 100 - 2 = 98$

The common difference is 5, the coefficient of $n$ .

The first four terms are 3, 8, 13, 18, and the 20th term is 98.

Question 3

Find an expression for the nth term of the sequence 6, 11, 16, 21, …

ShowHide solution

The common difference is $11 - 6 = 5$ , so compare with $5n$ : 5, 10, 15, 20.

The terms 6, 11, 16, 21 are each 1 more than $5n$ , so $T_{n} = 5n + 1$ .

Check: $T_{2} = 5 \times 2 + 1 = 11$ and $T_{4} = 5 \times 4 + 1 = 21$ .

The nth term is $5n + 1$ .

Question 4

Find an expression for the nth term of each sequence, and its 40th term.
(a) $-5$ , $-2$ , 1, 4, …
(b) 1.5, 2, 2.5, 3, …

ShowHide solution

(a) The common difference is $-2 - (-5) = -2 + 5 = 3$ , so compare with $3n$ : 3, 6, 9, 12.

The first term is $-5$ and $3n$ is 3: $-5 - 3 = -8$ , so the term is 8 less than $3n$ . So is the fourth: $4 - 12 = -8$ . So $T_{n} = 3n - 8$ .

Check: $T_{2} = 3 \times 2 - 8 = -2$ and $T_{4} = 3 \times 4 - 8 = 4$ .

$T_{40} = 3 \times 40 - 8 = 120 - 8 = 112$

(b) The common difference is $2 - 1.5 = 0.5$ , so compare with $0.5n$ : 0.5, 1, 1.5, 2.

Each term is 1 more than $0.5n$ , so $T_{n} = 0.5n + 1$ (which can also be written $\frac{n}{2} + 1$ ).

Check: $T_{3} = 0.5 \times 3 + 1 = 2.5$ and $T_{4} = 0.5 \times 4 + 1 = 3$ .

$T_{40} = 0.5 \times 40 + 1 = 20 + 1 = 21$

(a) $T_{n} = 3n - 8$ , and the 40th term is 112.
(b) $T_{n} = 0.5n + 1$ , and the 40th term is 21.

Question 5

Find an expression for the nth term of the sequence 35, 31, 27, 23, …, and find its 15th term.

ShowHide solution

The common difference is $31 - 35 = -4$ , so compare with $-4n$ : $-4$ , $-8$ , $-12$ , $-16$ .

The first term is $35 - (-4) = 39$ more than $-4n$ , and so is the second: $31 - (-8) = 39$ . So $T_{n} = -4n + 39 = 39 - 4n$ .

Check: $T_{2} = 39 - 8 = 31$ and $T_{4} = 39 - 16 = 23$ .

$T_{15} = 39 - 4 \times 15 = 39 - 60 = -21$

The nth term is $39 - 4n$ , and the 15th term is $-21$ .

Question 6

The sequence 5, 8, 11, 14, … has nth term $3n + 2$ . Is (a) 150, (b) 152 a term of the sequence? If it is, which term is it?

ShowHide solution

Every term is 2 more than a multiple of 3. So take away 2 and see whether a multiple of 3 is left.

(a) $150 - 2 = 148$ , and $148 \div 3 = 49$ remainder 1, so 148 is not a multiple of 3. (Indeed $T_{49} = 149$ and $T_{50} = 152$ .) 150 is not a term.

(b) $152 - 2 = 150 = 3 \times 50$ , so 152 is the term with $n = 50$ . Check: $3 \times 50 + 2 = 152$ .

(a) 150 is not a term.
(b) 152 is the 50th term.

Question 7

A sequence begins 5, 9, 13, … and ends at 201. How many terms does it have?

ShowHide solution

The common difference is 4, so compare with $4n$ : 4, 8, 12. Each term is 1 more, so $T_{n} = 4n + 1$ . Check: $T_{3} = 4 \times 3 + 1 = 13$ .

The last term is 201. $201 - 1 = 200 = 4 \times 50$ , so 201 is the 50th term. Check: $T_{50} = 4 \times 50 + 1 = 201$ .

The terms are $T_{1}$ , $T_{2}$ , …, $T_{50}$ , so there are 50 of them.

The sequence has 50 terms.

Question 8

Ravi makes a row of triangles with matchsticks. Figure $n$ has $n$ triangles.
(a) Draw Figure 4 and count its matchsticks.
(b) Find, in terms of $n$ , the number of matchsticks in Figure $n$ , and explain your answer from the drawing.
(c) How many matchsticks are there in Figure 30?
(d) Ravi has 80 matchsticks. Can he make a figure that uses all of them? What is the largest figure he can make?

ShowHide solution

(a) Figure 4 adds one more triangle, pointing down, to Figure 3. It needs 2 more matchsticks: $7 + 2 = 9$ .

(b) The numbers of matchsticks are 3, 5, 7, 9: the common difference is 2, so compare with $2n$ : 2, 4, 6, 8. Each is 1 more, so Figure $n$ has $2n + 1$ matchsticks.

From the drawing: there is one matchstick at the left end (the left side of the first triangle). Each triangle then needs 2 more matchsticks, because its left side is already there: for the first triangle it is the matchstick at the left end, and for every later triangle it is the side it shares with the triangle before it. So Figure $n$ has $1 + 2n = 2n + 1$ matchsticks.

(d) Take away the first matchstick: $80 - 1 = 79$ . The rest come in twos, and 79 is odd, so 80 is not $2n + 1$ for a whole number $n$ . He cannot use all 80.

Figure 39 uses $2 \times 39 + 1 = 79$ matchsticks and Figure 40 needs $2 \times 40 + 1 = 81$ , which is more than 80. So the largest figure is Figure 39, with 1 matchstick left over.

(a) Figure 4 has 9 matchsticks.
(b) $2n + 1$ matchsticks.
(c) 61 matchsticks.
(d) No. The largest figure he can make is Figure 39, which leaves 1 matchstick over.

Question 9

Hui Min arranges counters in crosses. Figure $n$ has a centre counter with $n$ counters on each of its four arms.
(a) Find, in terms of $n$ , the number of counters in Figure $n$ , and explain it from the drawing.
(b) How many counters are there in Figure 25?
(c) Is there a figure with exactly 200 counters?

ShowHide solution

(a) Counting: 5, 9, 13. The common difference is 4, so compare with $4n$ : 4, 8, 12. Each is 1 more, so Figure $n$ has $4n + 1$ counters.

From the drawing: there are 4 arms of $n$ counters each, which makes $4n$ , and 1 counter in the centre. So there are $4n + 1$ counters.

(b) $4 \times 25 + 1 = 101$

(c) Take away the centre counter: $200 - 1 = 199$ . The arms hold a multiple of 4, but $199 \div 4 = 49$ remainder 3. So no figure has exactly 200 counters. (Figure 49 has 197 and Figure 50 has 201.)

(a) $4n + 1$ counters.
(b) 101 counters.
(c) No figure has exactly 200 counters.

Question 10

Siti has $50 in her savings. At the end of each week she adds $8 of her pocket money, so at the end of weeks 1, 2 and 3 she has $58, $66 and $74.
(a) Write an expression for the amount she has at the end of week $n$ .
(b) How much does she have at the end of week 20?
(c) Will she ever have exactly $300 at the end of a week?
(d) At the end of which week does she first have more than $300?

ShowHide solution

(a) The amounts in dollars, 58, 66, 74, have a common difference of 8. Compare with $8n$ : 8, 16, 24. Each amount is 50 more, so at the end of week $n$ she has $(8n + 50)$ dollars. (The 50 is what she started with, and $8n$ is what she has added in $n$ weeks.)

(b) $8 \times 20 + 50 = 160 + 50 = 210$

(c) She has added $300 - 50 = 250$ dollars by then, and this must be a multiple of 8. $250 \div 8 = 31$ remainder 2, so it is not. She never has exactly $300 at the end of a week.

(d) From (c), $250 is 31 lots of $8 and $2 more, so look at weeks 31 and 32. At the end of week 31 she has $8 \times 31 + 50 = 248 + 50 = 298$ dollars, which is not more than $300. At the end of week 32 she has $8 \times 32 + 50 = 256 + 50 = 306$ dollars.

(a) $(8n + 50)$ dollars
(b) $210
(c) No.
(d) At the end of week 32 (she then has $306).

Question 11

(a) Write down the 15th square number, the 7th cube number and the 8th triangular number.
(b) Find an expression for the nth term of the sequence 0, 3, 8, 15, 24, …, and find its 10th term.

ShowHide solution

(a) $15^{2} = 225$ ; $7^{3} = 7 \times 7 \times 7 = 343$ ; $\frac{8 \times 9}{2} = \frac{72}{2} = 36$ (and $1 + 2 + … + 8 = 36$ ).

(b) The differences are 3, 5, 7, 9: no common difference. Compare with the square numbers 1, 4, 9, 16, 25: each term is 1 less. So $T_{n} = n^{2} - 1$ .

Check: $T_{3} = 3^{2} - 1 = 8$ and $T_{5} = 5^{2} - 1 = 24$ .

$T_{10} = 10^{2} - 1 = 100 - 1 = 99$

(a) 225, 343 and 36.
(b) $T_{n} = n^{2} - 1$ , and the 10th term is 99.

Question 12

The water tank for a school garden holds 500 litres when full. It is filled once, and then 15 litres are used each day, with no rain. At the end of days 1, 2 and 3 it holds 485, 470 and 455 litres.
(a) Write an expression for the number of litres left at the end of day $n$ .
(b) How much water is left at the end of day 20?
(c) Is the tank ever exactly empty at the end of a day? On which day does the water run out?

ShowHide solution

(a) The common difference is $470 - 485 = -15$ , so compare with $-15n$ : $-15$ , $-30$ , $-45$ . Each amount is 500 more: $485 - (-15) = 500$ . So $500 - 15n$ litres are left at the end of day $n$ . (500 litres at the start, less 15 litres for each of $n$ days.)

(b) $500 - 15 \times 20 = 500 - 300 = 200$

(c) For the tank to be empty, the water used, $15n$ litres, must be exactly 500 litres. $500 \div 15 = 33$ remainder 5, so 500 is not a multiple of 15: the tank is never exactly empty at the end of a day.

At the end of day 33 there are $500 - 15 \times 33 = 500 - 495 = 5$ litres left. Day 34 needs 15 litres, so the water runs out during day 34.

(a) $(500 - 15n)$ litres
(b) 200 litres
(c) No. 5 litres are left at the end of day 33, and the water runs out during day 34.

Key Points

A sequence is a list of numbers in order, made by a rule; each number is a term. $T_{n}$ is the nth term, and the term number $n$ is 1, 2, 3, ….
If the same number is added each time, it is the common difference: any term − the term before it. It is negative when the terms go down.
To find the nth term of a sequence with common difference $d$ : compare the terms with $dn$ , find the number that turns $dn$ into each term, and write $T_{n} = dn$ + that number.
For a sequence with a common difference, the coefficient of $n$ in the nth term is the common difference: 3, 7, 11, … has nth term $4n - 1$ , and 20, 17, 14, … has nth term $23 - 3n$ .
Always check the nth term on two terms of the sequence.
To find a term, substitute its term number for $n$ : the 50th term of $4n - 1$ is $4 \times 50 - 1 = 199$ .
To decide whether a number is a term, reason from the nth term: a term of $4n - 1$ is 1 less than a multiple of 4. The term number must be a positive whole number.
In a pattern of figures, the figure number is $n$ . Count, tabulate, find the nth term, and explain each part of it from the drawing.
Square numbers have nth term $n^{2}$ , cube numbers $n^{3}$ , and triangular numbers $\frac{n(n + 1)}{2}$ .