Maths · 2 Algebra

2.7 Sequences

Before you start

Substituting a whole number into an expression, including squares and negatives: if $n = 4$ , then $2 n^{2} - n = 28$ .
Solving a linear equation such as $6 n - 5 = 115$ , and a quadratic such as $n^{2} + n - 12 = 0$ by factorising.
Square numbers up to $15^{2}$ , cube numbers up to $5^{3}$ , and powers of 2 up to $2^{10} = 1024$ .
Index notation, including $2^{0} = 1$ and $3^{4} = 81$ .

Syllabus objectives covered (3)

Continue a given number sequence or pattern.
Recognise patterns in sequences, including the term-to-term rule, and relationships between different sequences.
Find and use the nth term of sequences.

Part 1 of 7

Sequences, terms, and continuing a pattern

A sequence is a list of numbers in a fixed order, made by a rule. Each number is a term, and its place in the list is its position, $n$ . The first term has $n = 1$ , the second has $n = 2$ , and so on.
The syllabus allows subscript notation. $T_{n}$ means the $n$ th term of a sequence called $T$ . For the sequence $5, 8, 11, 14, \dots$ you have $T_{1} = 5$ and $T_{3} = 11$ . The small number is the position; the value is the term.
To continue a sequence, find what happens from each term to the next. Write the gaps underneath: $5, 8, 11, 14$ has gaps $+ 3, + 3, + 3$ . Check the rule on every gap you are given, not just the first, and only then apply it. The next two terms are $17$ and $20$ .
A pattern of shapes is a sequence in disguise. Count the objects in each pattern, write the counts as a list, then continue the list. Here, Pattern 1 is one square of 4 matchsticks, Pattern 2 is two squares in a row (7 sticks), and Pattern 3 is three squares in a row (10 sticks).
graph diagram — not drawn yet
Three matchstick patterns side by side, labelled Pattern 1, Pattern 2 and Pattern 3. Pattern 1 is one square made of 4 sticks. Pattern 2 is two squares in a row sharing one vertical side, 7 sticks. Pattern 3 is three squares in a row, each neighbouring pair sharing one vertical side, 10 sticks. The number of sticks is written under each pattern.
Each new square shares a side with the square before it, so it adds 3 sticks, not 4.
Look at why the pattern grows, not only at the numbers. Each new square needs only 3 new sticks, because its left side is the right side of the square before. So Pattern 4 has $10 + 3 = 13$ sticks. When you know why it goes up by 3, you are sure of the next term instead of guessing it from three numbers.
The trap: a few terms on their own do not prove a rule. $1, 2, 4$ could continue $8, 16$ (multiply by 2) or $7, 11$ (gaps of $1, 2, 3, 4$ ). Use every term you are given to test your rule, and when a pattern is drawn, use the drawing to see how it is built.

Part 2 of 7

Term-to-term rules

A term-to-term rule tells you how to get from one term to the next. State it with the operation and the number: 'add 3', 'subtract 5', 'multiply by 2'. 'It goes up' is not a rule, because it does not let anyone make the next term.
A rule and a first term together fix the whole sequence: 'first term 5, add 3'. In subscript notation that is $T_{1} = 5$ and $T_{n + 1} = T_{n} + 3$ . Read $T_{n + 1}$ as 'the next term' and $T_{n}$ as 'this term'.
$T_{1} = 5, T_{n + 1} = T_{n} + 3$
If the same number is added each time, the gaps are all equal. They can be negative: $20, 17, 14, 11$ has the rule 'subtract 3', and the next term is $8$ .
If each term is the one before multiplied by the same number, divide a term by the term before it to find that number. For $3, 6, 12, 24$ , $24 \div 12 = 2$ , so the rule is 'multiply by 2'. For $80, 40, 20, 10$ , $10 \div 20 = \frac{1}{2}$ , so the rule is 'divide by 2'. This number is the common ratio.
Some rules use the two terms before. In $1, 1, 2, 3, 5, 8$ each term is the sum of the two before it, so the next term is $5 + 8 = 13$ . This is the Fibonacci sequence. A rule like this needs two starting terms, not one.
Some gaps change in a pattern of their own. $2, 3, 5, 8, 12$ has gaps $1, 2, 3, 4$ . Continue the row of gaps first (the next gap is $5$ ), then add it on: the next term is $17$ .
The weakness of a term-to-term rule is that to find the 100th term you must first find the 99th. To jump straight to any position you need a rule that links the position $n$ to the term. That is the $n$ th term, and it is the subject of the rest of this lesson.

Part 3 of 7

Sequences to recognise, and how sequences relate

Know these sequences on sight, with their $n$ th terms. The other sequences in this lesson are built from them.
$\begin{matrix} even numbers: 2, 4, 6, 8, \dots & 2 n \\ odd numbers: 1, 3, 5, 7, \dots & 2 n - 1 \\ square numbers: 1, 4, 9, 16, 25, \dots & n^{2} \\ cube numbers: 1, 8, 27, 64, 125, \dots & n^{3} \\ powers of 2: 2, 4, 8, 16, 32, \dots & 2^{n} \\ triangular numbers: 1, 3, 6, 10, 15, \dots & \frac{1}{2} n (n + 1) \end{matrix}$
Many sequences are one of these with something changed. Compare each term with the standard sequence in the same position. $2, 5, 10, 17, 26$ is $1, 4, 9, 16, 25$ with 1 added to each term, so $T_{n} = n^{2} + 1$ .
The change can be a multiplication. $3, 12, 27, 48$ is $1, 4, 9, 16$ with each term multiplied by 3, so $T_{n} = 3 n^{2}$ .
To find the relationship between two given sequences, write one above the other and look at each pair: is it a sum, a difference, or a multiple? Take $A$ : $2, 6, 12, 20, 30$ and $B$ : $1, 3, 6, 10, 15$ . Each term of $A$ is double the term of $B$ in the same position.
That relationship can tell you an $n$ th term. The terms of $A$ are $1 \times 2$ , $2 \times 3$ , $3 \times 4$ , $\dots$ , so $A_{n} = n (n + 1)$ . $B$ is half of $A$ , which is why the triangular numbers have $n$ th term $\frac{1}{2} n (n + 1)$ .
Sometimes the link is a shift in position, not a change in value. $4, 9, 16, 25$ are square numbers, but $4$ is the 2nd square, not the 1st. The $n$ th term of this sequence is the $(n + 1)$ th square, ${(n + 1)}^{2}$ . Check with $n = 1$ : ${(1 + 1)}^{2} = 4$ .
Two sequences can also be added term by term. $2^{n} + n$ gives $2 + 1, 4 + 2, 8 + 3, 16 + 4$ , which is $3, 6, 11, 20$ . When a sequence fits nothing you know, subtract a standard sequence from it and see whether what is left is simpler.

Part 4 of 7

The nth term of a linear sequence, and using it

A linear sequence has a constant gap, the common difference $d$ . Its $n$ th term always has the form $d n + c$ .
Here is why. $3 n$ gives the 3 times table, $3, 6, 9, 12$ , which goes up by 3. Every sequence that goes up by 3 is the 3 times table moved up or down. $5, 8, 11, 14$ is 2 more than $3, 6, 9, 12$ in every position, so $T_{n} = 3 n + 2$ .
$T_{n} = 3 n + 2$
A quick way to find $c$ : it is the term that would come before the first one, the 'zeroth term'. Work backwards one step: first term minus $d$ . For $5, 8, 11, 14$ that is $5 - 3 = 2$ .
For a decreasing sequence $d$ is negative, and the sign goes in front of $n$ . $20, 17, 14, 11$ has $d = - 3$ , and $c = 20 - (- 3) = 23$ , so $T_{n} = 23 - 3 n$ . Check with $n = 1$ and $n = 2$ : $23 - 3 = 20$ and $23 - 6 = 17$ . Always check two terms, because one term can match by accident.
A pattern gives the same formula from its structure. The matchstick squares had $4, 7, 10$ sticks: $d = 3$ and $c = 4 - 3 = 1$ , so $3 n + 1$ . In the drawing, that is 1 starting stick plus 3 sticks for each of the $n$ squares.
Use 1: find a term. Substitute the position. For $T_{n} = 3 n + 2$ , the 50th term is $3 (50) + 2 = 152$ .
Use 2: find a position. Set the $n$ th term equal to the value and solve. $3 n + 2 = 152$ gives $3 n = 150$ , so $n = 50$ : 152 is the 50th term.
Use 3: test whether a number is in the sequence. A position must be a positive whole number. $3 n + 2 = 100$ gives $3 n = 98$ and $n = 32.67$ (2 d.p.). That is not a whole number, so 100 is not a term: it falls between the 32nd term (98) and the 33rd term (101). Write that conclusion in words.
Use 4: the first term past a value. The first term over 100 needs $3 n + 2 > 100$ , so $n > 32.67$ , and the smallest whole $n$ is 33. The term is $3 (33) + 2 = 101$ .
These uses work for every kind of $n$ th term, not only linear ones. For $T_{n} = n^{2} + 1$ , the value 50 gives $n^{2} = 49$ , so $n = 7$ . Reject $n = - 7$ , because a position cannot be negative.

Part 5 of 7

Quadratic sequences: the difference method

If the first differences are not constant, find the differences of the differences, the second differences. If the second differences are constant, the sequence is quadratic: its $n$ th term is $a n^{2} + b n + c$ .
Here is where the method comes from. Substitute $n = 1, 2, 3, 4$ into $a n^{2} + b n + c$ , then take differences twice.
$\begin{matrix} terms: & a + b + c, 4 a + 2 b + c, 9 a + 3 b + c, 16 a + 4 b + c \\ 1st differences: & 3 a + b, 5 a + b, 7 a + b \\ 2nd differences: & 2 a, 2 a \end{matrix}$
So the second difference is always $2 a$ , which means ** $a$ is half the second difference**. The halving is the step the whole method depends on.
The method has four moves. (1) Find $a$ from the second difference. (2) Subtract $a n^{2}$ from each term. (3) What is left is linear, so find its $n$ th term $b n + c$ . (4) Put the parts together and check two terms.
Try it on $3, 4, 7, 12, 19$ . First differences: $1, 3, 5, 7$ . Second differences: $2, 2, 2$ . So $a = 2 \div 2 = 1$ , and the sequence starts with $n^{2}$ .
Subtract $n^{2}$ , which is $1, 4, 9, 16, 25$ , from each term: $3 - 1 = 2$ , $4 - 4 = 0$ , $7 - 9 = - 2$ , $12 - 16 = - 4$ , $19 - 25 = - 6$ . The remainder $2, 0, - 2, - 4, - 6$ is linear with $d = - 2$ and $c = 2 - (- 2) = 4$ , so it is $- 2 n + 4$ .
Put the parts together and check: $T_{n} = n^{2} - 2 n + 4$ . With $n = 1$ : $1 - 2 + 4 = 3$ . With $n = 5$ : $25 - 10 + 4 = 19$ . Both match.
$T_{n} = n^{2} - 2 n + 4$
The awkward case: an odd second difference gives a fraction. The triangular numbers $1, 3, 6, 10, 15$ have second difference 1, so $a = \frac{1}{2}$ . Subtracting $\frac{1}{2} n^{2}$ leaves $\frac{1}{2}, 1, \frac{3}{2}, 2, \dots$ , which is $\frac{1}{2} n$ . So $T_{n} = \frac{1}{2} n^{2} + \frac{1}{2} n$ , the same as $\frac{1}{2} n (n + 1)$ . A fraction here is correct, not a sign of error.
On a calculator paper you can check your answer with the graphic display calculator: enter it as a function, such as $y = x^{2} - 2 x + 4$ , and read its table at $x = 1, 2, 3, \dots$ . The table must show the terms you were given.

Part 6 of 7

Cubic sequences: one more row of differences

If the second differences are not constant either, take a third row. Constant third differences mean the sequence is cubic: its $n$ th term is $a n^{3} + b n^{2} + c n + d$ .
Find what $n^{3}$ alone gives. The cubes $1, 8, 27, 64, 125$ have first differences $7, 19, 37, 61$ , second differences $12, 18, 24$ , and third differences $6, 6$ . So $n^{3}$ has a third difference of 6, and $a n^{3}$ has a third difference of $6 a$ .
$a = \frac{third difference}{6}$
Then work as before. Subtract $a n^{3}$ from each term. What is left is quadratic, linear or constant, and you already have a method for each of those.
You need at least five terms to see two third differences and confirm that they are equal. With four terms you get only one third difference, and one number cannot show that a row is constant.
The pattern of divisors is worth remembering as one rule: a constant first difference is the coefficient of $n$ itself, a constant second difference is divided by 2 to give the coefficient of $n^{2}$ , and a constant third difference is divided by 6 to give the coefficient of $n^{3}$ .

Part 7 of 7

Exponential sequences and combinations

Some sequences never reach a constant row of differences. $3, 6, 12, 24, 48$ has differences $3, 6, 12, 24$ , which is the same doubling pattern again. A row of differences that copies the row above means the difference method will not finish. Test the ratios instead.
If every term divided by the term before gives the same number $r$ , the sequence is exponential (also called geometric). Its $n$ th term is the first term multiplied by $r^{n - 1}$ .
$T_{n} = T_{1} \times r^{n - 1}$
The power is $n - 1$ , not $n$ , because the first term has not been multiplied by $r$ at all, the second term has been multiplied once, and the $n$ th term has been multiplied $n - 1$ times. For $3, 6, 12, 24$ : $T_{n} = 3 \times 2^{n - 1}$ . Check: $T_{1} = 3 \times 2^{0} = 3$ .
Equivalent forms are equally correct. $3 \times 2^{n - 1} = 1.5 \times 2^{n}$ , because $2^{n - 1}$ is half of $2^{n}$ . Use whichever form you find, then check it with $n = 1$ .
A ratio less than 1 makes the terms smaller. $80, 40, 20, 10$ has $r = \frac{1}{2}$ , so $T_{n} = 80 \times {(\frac{1}{2})}^{n - 1}$ . Check: $T_{4} = 80 \times \frac{1}{8} = 10$ .
A combination is two sequences added term by term. If neither the differences nor the ratios settle, subtract the standard sequence that the terms grow like, and find the $n$ th term of what is left. Then add the two $n$ th terms together.
The clue to try a power: the differences themselves look like powers. If the differences are $2, 4, 8, 16$ , compare the terms with $2^{n}$ . If the differences are $6, 18, 54$ , compare the terms with a multiple of $3^{n}$ .

Worked examples

Example 1. The first four terms of a sequence are $1, 7, 13, 19$ . (a) Write down the next term and the term-to-term rule. (b) Find the $n$ th term. (c) Show that 100 is not a term of the sequence. (d) Find which term is equal to 115.

Show the working

Write the gaps: $6, 6, 6$ . Every gap is the same, so the rule is 'add 6' and the sequence is linear. The next term is $19 + 6 = 25$ .
For the $n$ th term, the difference 6 goes in front of $n$ , so it is $6 n + c$ . The zeroth term is $1 - 6 = - 5$ , so $c = - 5$ .
$T_{n} = 6 n - 5$
Check two terms: $6 (1) - 5 = 1$ and $6 (2) - 5 = 7$ . Both match.
For (c), set the $n$ th term equal to 100 and solve.
$6 n - 5 = 100 \Rightarrow 6 n = 105 \Rightarrow n = 17.5$
A position must be a whole number, and 17.5 is not. So 100 lies between the 17th term, $6 (17) - 5 = 97$ , and the 18th term, $6 (18) - 5 = 103$ . It is not in the sequence.
For (d), do the same with 115.
$6 n - 5 = 115 \Rightarrow 6 n = 120 \Rightarrow n = 20$
This time $n$ is a whole number, so 115 is a term, and $n$ tells you its position.

Answer: (a) $25$ ; the rule is 'add 6'. (b) $T_{n} = 6 n - 5$ . (c) $6 n - 5 = 100$ gives $n = 17.5$ , which is not a whole number, so 100 is not a term. (d) 115 is the 20th term.

Example 2. Find the $n$ th term of the sequence $1, 6, 17, 34, 57, \dots$ and use it to find the 10th term.

Show the working

First differences: $5, 11, 17, 23$ . They are not constant, so go one row further.
Second differences: $6, 6, 6$ . They are constant, so the sequence is quadratic.
Halve the second difference to get the coefficient of $n^{2}$ : $a = 6 \div 2 = 3$ , so the $n$ th term starts with $3 n^{2}$ .
Write out $3 n^{2}$ : $3, 12, 27, 48, 75$ . Subtract it from each term of the sequence.
$1 - 3 = - 2, 6 - 12 = - 6, 17 - 27 = - 10, 34 - 48 = - 14, 57 - 75 = - 18$
The remainder $- 2, - 6, - 10, - 14, - 18$ is linear with $d = - 4$ . Its zeroth term is $- 2 - (- 4) = 2$ , so it is $- 4 n + 2$ .
Put the parts together.
$T_{n} = 3 n^{2} - 4 n + 2$
Check two terms: $n = 1$ gives $3 - 4 + 2 = 1$ , and $n = 5$ gives $75 - 20 + 2 = 57$ . Both match.
Now use it. Substitute $n = 10$ .
$T_{10} = 3 (100) - 4 (10) + 2 = 300 - 40 + 2 = 262$

Answer: $T_{n} = 3 n^{2} - 4 n + 2$ , and the 10th term is $262$ .

Example 3. Find the $n$ th term of the sequence $2, 15, 52, 125, 246, \dots$

Show the working

First differences: $13, 37, 73, 121$ . Second differences: $24, 36, 48$ . Neither row is constant.
Third differences: $12, 12$ . They are constant, so the sequence is cubic. Five terms give exactly two third differences, which is enough to see that they match.
Divide the third difference by 6 to get the coefficient of $n^{3}$ : $a = 12 \div 6 = 2$ , so the $n$ th term starts with $2 n^{3}$ .
Write out $2 n^{3}$ : $2, 16, 54, 128, 250$ . Subtract it from each term.
$2 - 2 = 0, 15 - 16 = - 1, 52 - 54 = - 2, 125 - 128 = - 3, 246 - 250 = - 4$
The remainder $0, - 1, - 2, - 3, - 4$ is linear with $d = - 1$ , and its zeroth term is $0 - (- 1) = 1$ . So it is $- n + 1$ . Here the remainder came out linear; if it had been quadratic you would use the method for quadratics on it.
Put the parts together and check with $n = 3$ : $2 (27) - 3 + 1 = 52$ . It matches.
$T_{n} = 2 n^{3} - n + 1$

Answer: $T_{n} = 2 n^{3} - n + 1$ .

Example 4. The first five terms of a sequence are $3, 5, 9, 17, 33$ . (a) Find the $n$ th term. (b) Calculate the 10th term. (c) Find which term is equal to 257.

Show the working

First differences: $2, 4, 8, 16$ . Second differences: $2, 4, 8$ . The rows copy each other, so the difference method will never reach a constant row. That points to a power.
Test the ratios: $5 \div 3$ and $9 \div 5$ are not equal, so this is not a plain exponential sequence. It is close to one.
The differences are powers of 2, so compare each term with $2^{n}$ in the same position: $2, 4, 8, 16, 32$ . Each term is exactly 1 more.
$T_{n} = 2^{n} + 1$
Check two terms: $2^{1} + 1 = 3$ and $2^{5} + 1 = 33$ . Both match.
For (b), substitute $n = 10$ .
$T_{10} = 2^{10} + 1 = 1024 + 1 = 1025$
For (c), set the $n$ th term equal to 257. Then $2^{n} = 256$ , and $256 = 2^{8}$ , so $n = 8$ .

Answer: (a) $T_{n} = 2^{n} + 1$ . (b) $1025$ . (c) 257 is the 8th term.

Where people go wrong

Mistake: Writing the $n$ th term of $5, 8, 11, 14$ as $n + 3$ , because the rule is 'add 3'.

Instead: The common difference goes in front of $n$ : $T_{n} = 3 n + 2$ .

$n + 3$ gives $4, 5, 6, 7$ , which goes up by 1, not by 3. The term-to-term rule links neighbouring terms; the $n$ th term links the position to the term. They are different rules.

Mistake: Using the second difference as the coefficient of $n^{2}$ , so that a second difference of 6 gives $6 n^{2}$ .

Instead: Halve it: a second difference of 6 gives $3 n^{2}$ . For a cubic, divide the third difference by 6.

$a n^{2}$ has second difference $2 a$ . You can see this from $n^{2}$ alone: $1, 4, 9, 16$ has second difference 2, not 1.

Mistake: Writing $3 \times 2^{n}$ as the $n$ th term of $3, 6, 12, 24$ .

Instead: $T_{n} = 3 \times 2^{n - 1}$ , which is the same as $1.5 \times 2^{n}$ .

$3 \times 2^{n}$ gives 6 when $n = 1$ , so it starts one doubling too late. Substituting $n = 1$ shows the error at once.

Mistake: Saying that 100 is a term of $3 n + 2$ because the equation $3 n + 2 = 100$ has a solution.

Instead: The solution is $n = 32.67$ (2 d.p.). A position must be a positive whole number, so 100 is not a term.

There is a 32nd term and a 33rd term, but nothing between them. A decimal value of $n$ means the number falls between two terms.

Mistake: Choosing a rule from the first gap only, e.g. continuing $2, 3, 5, 8$ as $9, 10$ because $2$ to $3$ is 'add 1'.

Instead: Write every gap: $1, 2, 3$ . The next gaps are $4$ and $5$ , so the next terms are $12$ and $17$ .

A rule must produce every term you are given. 'Add 1' fails at the second gap, so it cannot be the rule.

Check yourself

State the term-to-term rule for $250, 50, 10, 2, \dots$ and write down the next two terms.
Answer
Each term divided by the one before is $\frac{1}{5}$ , so the rule is 'divide by 5'. The next two terms are $0.4$ and $0.08$ .
Write down the next two terms of $2, 2, 4, 6, 10, 16, \dots$ and describe the rule.
Answer
Each term is the sum of the two terms before it. The next two terms are $10 + 16 = 26$ and $16 + 26 = 42$ .
Sequence $A$ is $1, 4, 9, 16, 25$ and sequence $B$ is $0, 3, 8, 15, 24$ . Describe how $B$ is related to $A$ , and write down the $n$ th term of $B$ .
Answer
Each term of $B$ is 1 less than the term of $A$ in the same position. $A$ is the square numbers, so $B_{n} = n^{2} - 1$ .
Find the $n$ th term of $7, 4, 1, - 2, \dots$
Answer
$d = - 3$ , and the zeroth term is $7 - (- 3) = 10$ . So $T_{n} = 10 - 3 n$ . Check: $10 - 3 = 7$ and $10 - 6 = 4$ .
A sequence has $n$ th term $5 n - 2$ . Is 83 a term of the sequence? If it is, which term is it?
Answer
$5 n - 2 = 83$ gives $5 n = 85$ , so $n = 17$ . This is a whole number, so 83 is the 17th term.
Find the $n$ th term of $4, 10, 18, 28, 40, \dots$
Answer
First differences $6, 8, 10, 12$ ; second differences $2$ , so $a = 1$ . Subtracting $n^{2}$ leaves $3, 6, 9, 12, 15$ , which is $3 n$ . So $T_{n} = n^{2} + 3 n$ .
Find the $n$ th term of $2, 6, 18, 54, \dots$ and calculate the 6th term.
Answer
The common ratio is 3 and the first term is 2, so $T_{n} = 2 \times 3^{n - 1}$ . $T_{6} = 2 \times 3^{5} = 2 \times 243 = 486$ .
The third differences of a sequence are all 18. Write down the coefficient of $n^{3}$ in its $n$ th term.
Answer
$18 \div 6 = 3$ , so the $n$ th term starts with $3 n^{3}$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A sequence of patterns is made from sticks. Pattern 1 is one pentagon made from 5 sticks. Each new pattern adds one more pentagon to the right, sharing one side with the pentagon before it. Pattern 2 uses 9 sticks and Pattern 3 uses 13 sticks.

(a) Write down the number of sticks in Pattern 4. 1 mark No calculator

Mark scheme

B1 17 cao

(b) State the term-to-term rule for the number of sticks. 1 mark No calculator

Mark scheme

B1 add 4 oe, e.g. 'goes up by 4 each time'

(c) Find an expression, in terms of $n$ , for the number of sticks in Pattern $n$ . 2 marks No calculator

Mark scheme

B2 4n + 1 oe. If B2 is not scored, B1 for 4n + k with k ≠ 1, or for 4n seen

(d) Explain why no pattern in this sequence uses exactly 100 sticks. 2 marks No calculator

Mark scheme

M1 4n + 1 = 100 oe, or 99 ÷ 4 seen; or 97 and 101 found as the 24th and 25th terms; or 4n + 1 stated to be always odd; FT their (c) if linear
A1 n = 24.75 oe, with the statement that n is not a whole number so there is no such pattern; or 100 shown to lie between consecutive terms 97 and 101; or 100 is even but every term is odd, so 100 is not a term

Question 2. The first five terms of sequence $A$ are $3, 8, 15, 24, 35$ . The first five terms of sequence $B$ are $4, 9, 16, 25, 36$ .

(a) Find the $n$ th term of sequence $A$ . 3 marks No calculator

Mark scheme

M1 second differences of 2 found, or n² identified as part of the nth term
M1 their an² subtracted from the terms, e.g. 2, 4, 6, 8, 10; or three correct equations in a, b and c formed from a + b + c = 3, 4a + 2b + c = 8, 9a + 3b + c = 15
A1 n² + 2n oe, e.g. n(n + 2)

(b) Write down the $n$ th term of sequence $B$ . 1 mark No calculator

Mark scheme

B1 (n + 1)² oe, e.g. n² + 2n + 1

(c) Describe the relationship between the terms of sequence $A$ and sequence $B$ . 1 mark No calculator

Mark scheme

B1 each term of B is 1 more than the term of A in the same position oe

(d) Determine which term of sequence $A$ is equal to 288. 2 marks

Mark scheme

M1 n² + 2n = 288 oe, or (n + 1)² = 289; FT their (a)
A1 16 cao, with n = −18 rejected or not stated

Question 3. The first four terms of sequence $P$ are $4, 12, 36, 108$ . The first four terms of sequence $Q$ are $6, 16, 42, 116$ . Each term of $Q$ is found from the term of $P$ in the same position.

(a) Write down the next term of sequence $P$ . 1 mark No calculator

Mark scheme

B1 324 cao

(b) Find the $n$ th term of sequence $P$ . 2 marks No calculator

Mark scheme

M1 $k \times 3^{n - 1}$ or $k \times 3^{n}$ seen, any k ≠ 0
A1 $4 \times 3^{n - 1}$ oe, e.g. $\frac{4}{3} \times 3^{n}$

(c) Find the $n$ th term of sequence $Q$ . 2 marks No calculator

Mark scheme

M1 Q − P found as 2, 4, 6, 8 soi
A1 $4 \times 3^{n - 1} + 2 n$ oe; FT their (b) + 2n

(d) Calculate the 6th term of sequence $Q$ . 1 mark

Mark scheme

B1 984; FT their (c)

How this is examined

Finding the nth term of a quadratic sequence with the difference method: A 3-mark Find the nth term, with the first five terms of the sequence given as a list A row of constant second differences, then the coefficient of $n^{2}$ as half of it: a second difference of 2 gives $n^{2}$ , not $2 n^{2}$ . Then the terms minus $a n^{2}$ written out, e.g. $2, 4, 6, 8, 10$ , and the $n$ th term of that remainder. Those two steps are the method; the final answer is one expression, such as $n^{2} + 2 n$ , that gives the first term when $n = 1$ . For a cubic, the constant third difference divided by 6 gives the coefficient of $n^{3}$ .
Writing the nth term of a linear sequence, from a list of numbers or from a pattern of shapes: A 2-mark Find an expression for the nth term, after a 1-mark Write down of the next term or the next pattern The common difference as the number in front of $n$ , and the constant as the first term minus the difference: $5, 9, 13$ gives $4 n + 1$ . An expression with the right $n$ term and the wrong constant, such as $4 n + 5$ , is only partly correct. An answer of 'add 4' is a term-to-term rule, not an expression in $n$ .
Deciding whether a number is in a sequence, or which term it is: A 2-mark Explain or Show (that) a given number is not a term, or a Find of the position of a given term The $n$ th term set equal to the number and solved, e.g. $4 n + 1 = 100$ gives $n = 24.75$ . Then a conclusion in words: ' $n$ is not a whole number, so 100 is not a term'. A value of $n$ with no sentence does not explain anything. If $n$ is a whole number, that number is the position.
Finding the nth term of an exponential sequence, and of a sequence built from it: A 2-mark Find the nth term once a constant ratio is spotted, then a 2-mark Find for a second sequence that is the first one with something added A power of the ratio with the correct first term: $4 \times 3^{n - 1}$ for $4, 12, 36, 108$ , checked at $n = 1$ . For the combined sequence, the term-by-term subtraction written out, e.g. $2, 4, 6, 8$ , and its $n$ th term added on: $4 \times 3^{n - 1} + 2 n$ .
Continuing a sequence and stating its term-to-term rule: A 1-mark Write down of the next term or terms, and a 1-mark State of the rule Terms that follow from every gap given, not only the first. A rule that names the operation and the number: 'add 4', 'multiply by 3', 'add the two terms before'. 'It goes up' names neither, so it is not a rule.
Using one sequence to describe another: A 1-mark Write down the nth term of a second sequence, given a first sequence whose nth term you have found, or a 1-mark Describe of how the two are linked The link stated term by term and in the same position: 'each term of $B$ is 1 more than the term of $A$ in the same position'. Then the $n$ th term of $A$ changed to match, e.g. $n^{2} + 2 n + 1$ , or ${(n + 1)}^{2}$ if you recognise the squares.

Words to know

Sequence: A list of numbers in a fixed order, made by a rule.
Term: One number in a sequence. $T_{n}$ is the term in position $n$ of a sequence called $T$ .
Term-to-term rule: A rule that gives the next term from the term (or terms) before it, such as 'add 3' or 'multiply by 2'.
nth term: A rule that gives any term directly from its position $n$ , such as $T_{n} = 3 n + 2$ .
Linear sequence: A sequence with a constant first difference. Its $n$ th term has the form $d n + c$ .
Common difference: The constant amount added from one term to the next in a linear sequence.
Quadratic sequence: A sequence with constant second differences. Its $n$ th term has the form $a n^{2} + b n + c$ , where $a$ is half the second difference.
Cubic sequence: A sequence with constant third differences. The coefficient of $n^{3}$ in its $n$ th term is the third difference divided by 6.
Exponential sequence: A sequence where each term is the one before multiplied by the same number. Its $n$ th term is $T_{1} \times r^{n - 1}$ .
Common ratio: The number $r$ that each term of an exponential sequence is multiplied by to give the next term.
Triangular numbers: The sequence $1, 3, 6, 10, 15, \dots$ , with $n$ th term $\frac{1}{2} n (n + 1)$ .

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