Sequences (Higher)

What you'll learn

How to describe number patterns with equal gaps.
How to find a far-away term without writing out the whole list.
How to add lots of terms quickly using a formula.
How to solve Grade 8/9 algebra questions involving sequences.

Sequences, terms, and equal gaps

A number pattern is usually written in order, for example:

…5,\ 8,\ 11,\ 14,\ 17,\ \ldots

The dots mean the pattern continues.

Definition

Sequence and term

A sequence is an ordered list of numbers. Each number in the list is called a term. The first term is the number in position 1, the second term is the number in position 2, and so on.

We write $T_n$ for the term in position $n$ .

An important type of sequence is one where the same amount is added each time.

Definition

Arithmetic sequence

An arithmetic sequence is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, written as $d$ .

If the first term is $a$ , then the nth term is

T_n=a+(n-1)d

The reason for the $n - 1$ is important: to get from the first term to the nth term, you make one fewer jump than the term number.

Arithmetic sequence equal jumps and paired series formula

Example

Finding a later term

A sequence begins 6, 10, 14, 18, 22, ... Find the 40th term.

Identify the first term and the common difference:
$a=6,d=10−6=4a=6,\qquad d=10-6=4$
Use the nth-term formula:
$T_n=a+(n-1)d$
Substitute $n = 40$ :
$T40=6+(40−1)×4T_{40}=6+(40-1)\times 4$
Simplify:
$T40=6+39×4=162T_{40}=6+39\times 4=162$
The 40th term is 162.

Common Mistake

Using too many jumps

The 40th term is not found by adding 40 lots of the common difference. From the first term to the 40th term there are 39 jumps, so you add $39 d$ .

Adding an arithmetic sequence: arithmetic series

Sometimes you are asked to add the first 50, 100, or even more terms. Writing them all out would be slow, so we use a formula.

Definition

Arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence.

We write $S_n$ for the sum of the first $n$ terms.

There are two main versions of the formula.

If you know the first term $a$ , the last term $l$ , and the number of terms $n$ :

Sn=n2(a+l)S_n=\frac{n}{2}(a+l)

If you know the first term $a$ , the common difference $d$ , and the number of terms $n$ :

Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)

Key Idea

Why the series formula works

Pair the first and last terms, then the second and second-last terms. Each pair has the same total, so the sum is based on an average term multiplied by the number of terms.

Example

Finding the sum of the first 100 terms

The first five terms of an arithmetic sequence are 4, 7, 10, 13, 16. Find the sum of the first 100 terms.

Identify the first term, common difference, and number of terms:
$a=4,d=3,n=100a=4,\qquad d=3,\qquad n=100$
Choose the series formula using $a$ , $d$ , and $n$ :
$Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)$
Substitute the values:
$S100=1002(2×4+(100−1)×3)S_{100}=\frac{100}{2}\left(2\times 4+(100-1)\times 3\right)$
Simplify inside the bracket first:
$S_{100}=50(8+297)$
Calculate the sum:
$S100=50×305=15250S_{100}=50\times 305=15250$

Tip

Which sum formula should you use?

Use $Sn=n2(a+l)S_n=\frac{n}{2}(a+l)$ when you know the first and last terms. Use $Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)$ when you know the first term and common difference.

Working backwards from two given terms

In harder questions, you may not be given the first term or common difference directly. Instead, you might be told two terms from the same arithmetic sequence.

The method is to turn each statement into an equation using

T_n=a+(n-1)d

Then solve the simultaneous equations. Simultaneous equations are equations that must both be true at the same time.

Example

Using two known terms to find a sum

The 6th term of an arithmetic sequence is 23. The 14th term is 55. Find the sum of the first 50 terms.

Write an equation for the 6th term:
$a + 5 d = 23$
Write an equation for the 14th term:
$a + 13 d = 55$
Subtract the first equation from the second equation:
$8 d = 32$
Find the common difference:
$d = 4$
Substitute $d = 4$ into $a + 5 d = 23$ :
$a+5×4=23a+5\times 4=23$
Find the first term:
$a = 3$
Use the arithmetic series formula:
$S50=502(2×3+(50−1)×4)S_{50}=\frac{50}{2}\left(2\times 3+(50-1)\times 4\right)$
Simplify:
$S_{50}=25(6+196)=5050$

Common Mistake

Mixing up term position and term value

If the 6th term is 23, the position is 6 and the value is 23. In the formula, the position decides how many lots of $d$ you have.

When two sums are compared

A Grade 8/9 question may compare two sums, for example saying that one sum is several times another sum. You may feel there is not enough information, but the algebra often makes the unknowns cancel.

The key is to write both sums using the same $a$ and $d$ , because they come from the same arithmetic sequence.

Example

Comparing two sums

The sum of the first 56 terms of an arithmetic series is 4 times the sum of the first 42 terms of the same series. Find the sum of the first 35 terms.

Write a formula for $S_{56}$ :
$S56=562(2a+55d)=28(2a+55d)S_{56}=\frac{56}{2}\left(2a+55d\right)=28(2a+55d)$
Write a formula for $S_{42}$ :
$S42=422(2a+41d)=21(2a+41d)S_{42}=\frac{42}{2}\left(2a+41d\right)=21(2a+41d)$
Use the information that $S_{56}$ is 4 times $S_{42}$ :
$28(2a+55d)=4×21(2a+41d)28(2a+55d)=4\times 21(2a+41d)$
Simplify the right-hand side:
$28 (2 a + 55 d) = 84 (2 a + 41 d)$
Divide both sides by 28:
$2 a + 55 d = 3 (2 a + 41 d)$
Expand and solve for $a$ in terms of $d$ :
$2a+55d=6a+123d−68d=4aa=−17d\begin{aligned} 2a+55d&=6a+123d\\ -68d&=4a\\ a&=-17d \end{aligned}$
Now find $S_{35}$ :
$S35=352(2a+34d)S_{35}=\frac{35}{2}\left(2a+34d\right)$
Substitute $a = - 17 d$ :
$S35=352(−34d+34d)=0S_{35}=\frac{35}{2}(-34d+34d)=0$
The sum of the first 35 terms is zero.

Common Mistake

A zero sum can be correct

A sum can be zero if negative and positive terms cancel each other out. Do not invent a value for $d$ unless the question gives enough information to find it.

Algebraic consecutive terms

Sometimes the terms are given as algebraic expressions. You must use the fact that the gaps are equal.

Definition

Consecutive terms

Consecutive terms are terms next to each other in the sequence, with no missing terms between them.

For three consecutive terms in an arithmetic sequence:

term\text{second term}-\text{first term}=\text{third term}-\text{second term}

Example

Proving the common difference

The expressions $2 x + 7$ , $6 x - 4$ , and $7 x$ are three consecutive terms of an arithmetic sequence. Prove that the common difference is 9.

Set the two consecutive differences equal:
$(6 x - 4) - (2 x + 7) = 7 x - (6 x - 4)$
Simplify both sides:
$4 x - 11 = x + 4$
Solve for $x$ :
$3 x = 15$
Therefore:
$x = 5$
Substitute $x = 5$ into one of the differences:
$(6 x - 4) - (2 x + 7) = 4 x - 11$
Calculate the common difference:
$4×5−11=94\times 5-11=9$
The common difference is 9, as required.

Tip

Proof questions need a conclusion

In a “prove” question, do not stop as soon as you find $x$ . You must calculate the common difference and write a clear final statement.

Exam technique

In the exam

Define $a$ , $d$ , and $n$ before substituting into formulas.
Use $T_n=a+(n-1)d$ for an individual term, and $Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)$ for a sum.
For algebraic proof questions, compare the two consecutive differences, solve for the variable, then calculate the common difference.

Self review

Check yourself

What is the difference between $T_n$ and $S_n$ ?
Why does the nth-term formula use $n - 1$ lots of the common difference?
If three algebraic expressions are consecutive arithmetic terms, what equation should you write first?

Sequences (Higher)

What you'll learn

How to describe number patterns with equal gaps.
How to find a far-away term without writing out the whole list.
How to add lots of terms quickly using a formula.
How to solve Grade 8/9 algebra questions involving sequences.

Sequences, terms, and equal gaps

A number pattern is usually written in order, for example:

…5,\ 8,\ 11,\ 14,\ 17,\ \ldots

The dots mean the pattern continues.

Definition

Sequence and term

A sequence is an ordered list of numbers. Each number in the list is called a term. The first term is the number in position 1, the second term is the number in position 2, and so on.

We write $T_n$ for the term in position $n$ .

An important type of sequence is one where the same amount is added each time.

Definition

Arithmetic sequence

An arithmetic sequence is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, written as $d$ .

If the first term is $a$ , then the nth term is

T_n=a+(n-1)d

The reason for the $n - 1$ is important: to get from the first term to the nth term, you make one fewer jump than the term number.

Arithmetic sequence equal jumps and paired series formula

Example

Finding a later term

A sequence begins 6, 10, 14, 18, 22, ... Find the 40th term.

Identify the first term and the common difference:
$a=6,d=10−6=4a=6,\qquad d=10-6=4$
Use the nth-term formula:
$T_n=a+(n-1)d$
Substitute $n = 40$ :
$T40=6+(40−1)×4T_{40}=6+(40-1)\times 4$
Simplify:
$T40=6+39×4=162T_{40}=6+39\times 4=162$
The 40th term is 162.

Common Mistake

Using too many jumps

The 40th term is not found by adding 40 lots of the common difference. From the first term to the 40th term there are 39 jumps, so you add $39 d$ .

Adding an arithmetic sequence: arithmetic series

Sometimes you are asked to add the first 50, 100, or even more terms. Writing them all out would be slow, so we use a formula.

Definition

Arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence.

We write $S_n$ for the sum of the first $n$ terms.

There are two main versions of the formula.

If you know the first term $a$ , the last term $l$ , and the number of terms $n$ :

Sn=n2(a+l)S_n=\frac{n}{2}(a+l)

If you know the first term $a$ , the common difference $d$ , and the number of terms $n$ :

Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)

Key Idea

Why the series formula works

Pair the first and last terms, then the second and second-last terms. Each pair has the same total, so the sum is based on an average term multiplied by the number of terms.

Example

Finding the sum of the first 100 terms

The first five terms of an arithmetic sequence are 4, 7, 10, 13, 16. Find the sum of the first 100 terms.

Identify the first term, common difference, and number of terms:
$a=4,d=3,n=100a=4,\qquad d=3,\qquad n=100$
Choose the series formula using $a$ , $d$ , and $n$ :
$Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)$
Substitute the values:
$S100=1002(2×4+(100−1)×3)S_{100}=\frac{100}{2}\left(2\times 4+(100-1)\times 3\right)$
Simplify inside the bracket first:
$S_{100}=50(8+297)$
Calculate the sum:
$S100=50×305=15250S_{100}=50\times 305=15250$

Tip

Which sum formula should you use?

Use $Sn=n2(a+l)S_n=\frac{n}{2}(a+l)$ when you know the first and last terms. Use $Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)$ when you know the first term and common difference.

Working backwards from two given terms

In harder questions, you may not be given the first term or common difference directly. Instead, you might be told two terms from the same arithmetic sequence.

The method is to turn each statement into an equation using

T_n=a+(n-1)d

Then solve the simultaneous equations. Simultaneous equations are equations that must both be true at the same time.

Example

Using two known terms to find a sum

The 6th term of an arithmetic sequence is 23. The 14th term is 55. Find the sum of the first 50 terms.

Write an equation for the 6th term:
$a + 5 d = 23$
Write an equation for the 14th term:
$a + 13 d = 55$
Subtract the first equation from the second equation:
$8 d = 32$
Find the common difference:
$d = 4$
Substitute $d = 4$ into $a + 5 d = 23$ :
$a+5×4=23a+5\times 4=23$
Find the first term:
$a = 3$
Use the arithmetic series formula:
$S50=502(2×3+(50−1)×4)S_{50}=\frac{50}{2}\left(2\times 3+(50-1)\times 4\right)$
Simplify:
$S_{50}=25(6+196)=5050$

Common Mistake

Mixing up term position and term value

If the 6th term is 23, the position is 6 and the value is 23. In the formula, the position decides how many lots of $d$ you have.

When two sums are compared

A Grade 8/9 question may compare two sums, for example saying that one sum is several times another sum. You may feel there is not enough information, but the algebra often makes the unknowns cancel.

The key is to write both sums using the same $a$ and $d$ , because they come from the same arithmetic sequence.

Example

Comparing two sums

The sum of the first 56 terms of an arithmetic series is 4 times the sum of the first 42 terms of the same series. Find the sum of the first 35 terms.

Write a formula for $S_{56}$ :
$S56=562(2a+55d)=28(2a+55d)S_{56}=\frac{56}{2}\left(2a+55d\right)=28(2a+55d)$
Write a formula for $S_{42}$ :
$S42=422(2a+41d)=21(2a+41d)S_{42}=\frac{42}{2}\left(2a+41d\right)=21(2a+41d)$
Use the information that $S_{56}$ is 4 times $S_{42}$ :
$28(2a+55d)=4×21(2a+41d)28(2a+55d)=4\times 21(2a+41d)$
Simplify the right-hand side:
$28 (2 a + 55 d) = 84 (2 a + 41 d)$
Divide both sides by 28:
$2 a + 55 d = 3 (2 a + 41 d)$
Expand and solve for $a$ in terms of $d$ :
$2a+55d=6a+123d−68d=4aa=−17d\begin{aligned} 2a+55d&=6a+123d\\ -68d&=4a\\ a&=-17d \end{aligned}$
Now find $S_{35}$ :
$S35=352(2a+34d)S_{35}=\frac{35}{2}\left(2a+34d\right)$
Substitute $a = - 17 d$ :
$S35=352(−34d+34d)=0S_{35}=\frac{35}{2}(-34d+34d)=0$
The sum of the first 35 terms is zero.

Common Mistake

A zero sum can be correct

A sum can be zero if negative and positive terms cancel each other out. Do not invent a value for $d$ unless the question gives enough information to find it.

Algebraic consecutive terms

Sometimes the terms are given as algebraic expressions. You must use the fact that the gaps are equal.

Definition

Consecutive terms

Consecutive terms are terms next to each other in the sequence, with no missing terms between them.

For three consecutive terms in an arithmetic sequence:

term\text{second term}-\text{first term}=\text{third term}-\text{second term}

Example

Proving the common difference

The expressions $2 x + 7$ , $6 x - 4$ , and $7 x$ are three consecutive terms of an arithmetic sequence. Prove that the common difference is 9.

Set the two consecutive differences equal:
$(6 x - 4) - (2 x + 7) = 7 x - (6 x - 4)$
Simplify both sides:
$4 x - 11 = x + 4$
Solve for $x$ :
$3 x = 15$
Therefore:
$x = 5$
Substitute $x = 5$ into one of the differences:
$(6 x - 4) - (2 x + 7) = 4 x - 11$
Calculate the common difference:
$4×5−11=94\times 5-11=9$
The common difference is 9, as required.

Tip

Proof questions need a conclusion

In a “prove” question, do not stop as soon as you find $x$ . You must calculate the common difference and write a clear final statement.

Exam technique

In the exam

Define $a$ , $d$ , and $n$ before substituting into formulas.
Use $T_n=a+(n-1)d$ for an individual term, and $Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\left(2a+(n-1)d\right)$ for a sum.
For algebraic proof questions, compare the two consecutive differences, solve for the variable, then calculate the common difference.

Self review

Check yourself

What is the difference between $T_n$ and $S_n$ ?
Why does the nth-term formula use $n - 1$ lots of the common difference?
If three algebraic expressions are consecutive arithmetic terms, what equation should you write first?

Sequences (Higher)

What you'll learn

Sequences, terms, and equal gaps

Adding an arithmetic sequence: arithmetic series

Working backwards from two given terms

When two sums are compared

Algebraic consecutive terms

Recap questions

You've reached the end

Sequences (Higher)

What you'll learn

Sequences, terms, and equal gaps

Adding an arithmetic sequence: arithmetic series

Working backwards from two given terms

When two sums are compared

Algebraic consecutive terms

Recap questions

You've reached the end

Sequences (Higher)

What you'll learn

Sequences, terms, and equal gaps

Adding an arithmetic sequence: arithmetic series

Working backwards from two given terms

When two sums are compared

Algebraic consecutive terms

Recap questions

You've reached the end

Sequences (Higher) Revision Guide

Sequences (Higher)

What you'll learn

Sequences, terms, and equal gaps

Adding an arithmetic sequence: arithmetic series

Working backwards from two given terms

When two sums are compared

Algebraic consecutive terms

Recap questions

You've reached the end

Sequences (Higher) Revision Guide