Arithmetic Sequences and Sums

GCSELevel 6-7Edexcel iGCSE

Arithmetic Sequences and Sums Revision

Arithmetic Sequences and Sums

An arithmetic sequence is a list of numbers, called terms. There is a common difference between consecutive terms, to for each new term the same number is added to or subtracted from the previous term. The numbers in the sequence can then be generated by a rule. We call this the $n$ th term rule.

This rule allows us to find a value at the $n$ th position in the sequence, which we call the $n$ th term.

We are also able to use the $n$ th term rule to take a sum of the first $n$ terms of a sequence.

Finding the $n$ th term

An arithmetic sequence is a sequence where every term has the same difference $d$ , between it and the next term. We can use the $n$ th term rule to generate any value in an arithmetic sequence. The rule is

The $n$ th term $= a + (n-1)d$ ,

where $d$ is the common difference, and $a$ is the very first term in the sequence.

Example: Find the $10$ th term in the following sequence,

Step 1: Find the first term ( $a$ )

In this sequence the first term is given as $3$ .

Step 2: Find the common difference ( $d$ )

The common difference is the difference between two consecutive terms. In this case we can chose the first and second terms to find $d$ ,

$d = 2\text{nd term} - 1\text{st term} = 7 - 3 = 4$ .

Step 3: Use the formula

$a = 3$ , $d = 4$ , so the $10$ th term is

$3 + (10 - 1) \times 4 = 3 + 9\times 4 = 3 + 36 = 39$

So the $10$ th term in the sequence is $39$ .

Finding the Sum of a Sequence

We can use a formula to find the sum of a sequence from the first term to the $n$ th term, where $n$ is the number of terms. We denote this sum as $S_n$

There are two formulas that can be used, although they are essentially the same.

The first formula:

$S_n = \dfrac{n}{2} (2a + (n-1)d)$ ,

Where $a$ is the first term in the sequence, $n$ is the number of terms and $d$ is the common difference.

From this first formula can find the second formula. We can deduce that $2a + (n-1)d = a + (a + (n-1)d)$ and by the $n$ th term rule we know that the $n$ th term $= a + (n-1)d$ . We can call the $n$ th term $a_n$ and the first term $a = a_1$ .

So $2a + (n-1)d = a_1 + a_n$ , which can be substituted back into the first formula.

The second formula:

$S_n = \dfrac{n}{2} (a_1 + a_n)$

Where $a_1$ is the first term, $a_n$ is the final term, and $n$ is the number of terms.

Example 1: Finding the $13$ th Term

Consider the sequence:

$1, 4, 7, 10, \dots$

Find the $13$ th term in the sequence.

[3 marks]

Firstly we can find the common difference:

$2$ nd term $- 1$ st term $= 4 - 1 = 3$
$3$ nd term $- 2$ st term $= 7 - 4 = 3$
$4$ nd term $- 3$ st term $= 10 - 7 = 3$

The common difference $(d) = 3$ .

The first term $(a) =1$ .

This gives us the $n$ th term rule of

$a_n = 1 + (n-1)\times 3$ .

For the $13$ th term,

$a_{13} = 1 + (13-1) \times 3$
$a_{13} = 1 + 12 \times 3$ ,
$a_{13} = 1 + 36$ ,
$a_{13} = 37$ ,

The $13$ th term of the sequence is $37$ .

Example 2: Finding the sum to the $1000$ th term

Consider the sequence:

$1, 2, 3, 4, \dots$

Find the sum of the first $1000$ terms in the sequence.

[3 marks]

The common difference, $d = 1$ . The first term, $a = 1$

We can use the first formula for sum of an arithmetic sequence to find the sum up to $n =1000$ .

$S_{1000} = \dfrac{1000}{2} (2 \times 1 + (1000-1) \times 1)$

$S_{1000} = \dfrac{1000}{2} (2 + (999))$

$S_{1000} = \dfrac{1000}{2} (1001)$

$S_{1000} = 500 \times (1001)$

$S_{1000} = 500,500$

Example 3: Determining if a value is a term in a sequence

Consider the sequence:

$15, 21, 27, 33, \dots$

determine whether or not the value $99$ is a term in the sequence.

[3 marks]

Firstly we must find the common difference:

$2$ nd term $- 1$ st term $= 21 - 15 = 6$
$3$ rd term $- 2$ nd term $= 27 - 21 = 6$
$4$ th term $- 3$ rd term $= 33 - 27 = 6$

The common difference $(d) = 6$ .

Therefore the $n$ th term rule is $15 + (n - 1) \times 6$

Now we can equate that to $99$ to see if it is a term,

$99 = 15 + (n-1) \times 6$ ,
$84 = (n-1) \times 6$ ,
$84 \div 6 = n - 1$ ,
$14 = n - 1$ ,
$n = 15$ ,

So yes, $99$ is the $15$ th term in the sequence.

Example 4: Using the Second Summation Formula

An arithmetic sequence has two known terms. The first term is $a_1 = 7$ and the $43$ rd term is $a_{43} = 343$ .

Find the $n$ th term rule and the sum of the sequence up to the $43$ rd term.

[4 marks]

The $n$ th term rule:

$a_n = a_1 + (n-1)d$

Let $n = 43$ then,

$a_{43} = 343 = 7 + (43 - 1)d$ ,
$343 - 7 = 42d$ ,
$336 = 42d$ ,
$336 \div 42 = d$ ,
$d = 8$ .

The $n$ th term rule then is:

$a_n = 7 + (n-1)\times 8$

The sum to the $43$ rd term is:

$S_{43} = \dfrac{43}{2} \times (a_1 + a_{43})$

$S_{43} = \dfrac{43}{2} \times (7 + 343)$

$S_{43} = \dfrac{43}{2} \times 350$

$S_{43} = 7525$

Arithmetic Sequences and Sums Example Questions

Question 1: The first $5$ terms of an arithmetic sequence are

$3,\,\,8,\,\,13,\,\,18,\,\,23$

Find the formula for the $n^{th}$ term of this sequence.

[2 marks]

The first term, $a = 3$ . We must now find the common difference.

$2$ nd term $- 1$ st term $= 8 - 3 = 5$ .

The common difference, $d = 5$ .

So the $n$ th term formula is

$3 + (n-1) \times 5=5n-2$

Question 2: The first $5$ terms of an arithmetic sequence are

$-6,\,\,-4,\,\,-2,\,\,0,\,\,2$

Find the sum of the first $30$ terms of this sequence.

[4 marks]

The first term, $a = -6$ . We must now find the common difference.

$2$ nd term $- 1$ st term $= -4 - (-6) = 2$ .

The common difference, $d = 2$ .

Summation formula:

$S_n = \dfrac{n}{2} (2a + (n-1)d)$ ,

so,

$S_{30} = \dfrac{30}{2} (2 \times (-6) + (30 - 1) \times 2)$

$S_{30} = 15 (-12 + (29) \times 2)$

$S_{30} = 15 (-12 + 58) = 15 \times 46 = 690$

Question 2: A sequence is defined by the formula $1080 + (n-1)(-40)$

a) Work out the first $5$ terms of this sequence.

[2 marks]

b) Determine whether or not $-140$ is in the sequence.

[2 marks]

a) To generate the first $5$ terms of this sequence, we will substitute $n=1, 2, 3, 4, 5$ into the formula given.

$1$ st $= 1080 + ((1)-1)(-40) = 1080$
$2$ nd $= 1080 + ((2)-1)(-40) = 1040$
$3$ rd $= 1080 + ((3)-1)(-40) = 1000$
$4$ th $= 1080 + ((4)-1)(-40) = 960$
$5$ th $= 1080 + ((5)-1)(-40) = 920$

So, the first $5$ terms are $1080$ , $1040$ , $1000$ , $960$ , and $920$

b) We can equate $-140$ to the $n$ th term rule, to find it’s position in the sequence. If it’s position is a whole number, it must be in the sequence, if not it can’t be.

$-140 = 1080 + (n-1)(-40)$
$-1220 = (n-1)(-40)$
$-1220 \div -40 = (n-1)$
$-1220 \div (-40) = 30.5 = (n-1)$
$n = 31.5$

As $n$ is not an integer, $-140$ is not in the sequence.

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