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3. Growth & Decay in Sequences
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Year 12

3.02 Arithmetic sequences

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Arithmetic sequences

A sequence in which each term changes from the last by adding a constant amount is called an arithmetic sequence. We refer to the constant the terms are changing by as the common difference, which will result from subtracting any two successive terms \left(t_{n+1}-t_n\right).

The progression -3,\,5,\,13,\,21,\,\ldots is an arithmetic progression with a common difference of 8. On the other hand, the progression 1,\,10,\,100,\,1000,\,\ldots is not arithmetic because the difference between each term is not constant.

We denote the first term by the letter a and the common difference by the letter d. Since, t_2=t_1+d, t_3=t_2+d and so on, we can write any arithmetic sequence as the recurrence relation: t_n=t_{n-1}+d,t_1=a.

If we consider the recurrence relation t_n=t_{n-1}+2,t_1=5, this recurrence relation starts with 5 and we add 2 to find the next term, therefore the sequence is 5,\,7,\,9,\,11,\,13,\,15....

We can also find an explicit formula in terms of a and d, this is useful for finding the nth term without listing the sequence.

nt_{n}\text{Pattern}
1-3-3
25-3+8
313-3+2\times8
421-3+3\times8
...
nt_n-3+(n-1)\times8

This table shows the pattern for an explicit formula. For the sequence -3,\,5,\,13,\,21,\,\ldots , we have starting term of -3 and a common difference of 8, that is a=3 and d=8.

From the pattern we could guess that the tenth term is t_{10}=-3+9\times 8=69 and the one hundredth term is t_{100}=789=-3+99\times 8. The explicit formula for the nth term is t_n=-3+(n-1)\times 8.

We could create a similar table for the arithmetic progression with starting value a and common difference d and we would observe the same pattern. Hence, generating the explicit rule for any arithmetic sequence is t_n=a+\left(n-1\right)d

For any arithmetic sequence with starting value a and common difference d, we can express it in either of the following two forms:

  • Recursive form is a way to express any term in relation to the previous term: t_n=t_{n-1}+d, where t_1=a

  • Explicit form is a way to express any term in relation to the term number: t_n=a+\left(n-1\right)d

Examples

Example 1

Determine which of the following sequences is an arithmetic progression.

a
A
3,\, 0,\, - 3 ,\, - 6 ,\, \ldots
B
1,\, 2, \,3, \,5,\, 8,\, 13,\, \ldots
C
3,\, 3^{3}, \,3^{6},\, 3^{9},\, \ldots
D
4, \, - 4 ,\, 4, \, - 4 ,\, \ldots
Worked Solution
Create a strategy

Check whether each sequence is adding or subtracting the same amount each time to get to the next term.

Apply the idea

For Option A, 3 is being subtracted each time: 3-3=0, \, 0-3=-3, \ldots.

For Option B, 1 is added the first two times but then 2 is added, then 3 is added and so on: 1+1=2, \, 3+2=5, \,5+3=8,\ldots.

For Option C, the sequence is made by adding 3 to the power of 3, but not to the actual terms.

For Option D, the term first decreases by 8 but then increases by 8: 4-8=-4, \, -4+8=4, \ldots.

The correct answer is Option A as we subtract the same amount each time to get to the next term.

b

What is the common difference of this progression?3,\, 0,\, - 3 ,\, - 6 ,\, \ldots

Worked Solution
Create a strategy

Find the difference between the second term and the first term.

Apply the idea

Notice that the terms are decreasing. This means the common difference will be negative.

\displaystyle \text{Common difference}\displaystyle =\displaystyle 0-3Subtract 3 from 0
\displaystyle =\displaystyle -3Evaluate

Example 2

The nth term of a sequence is defined by: T_n = 15+5\left(n-1\right).

a

Determine a, the first term in the arithmetic progression.

Worked Solution
Create a strategy

Substitute n=1 into the rule.

Apply the idea
\displaystyle T_1\displaystyle =\displaystyle 15+5\times\left(1-1\right)Substitute n=1
\displaystyle a\displaystyle =\displaystyle 15+5\times 0Evaluate the subtraction
\displaystyle =\displaystyle 15Evaluate
b

Determine d, the common difference.

Worked Solution
Create a strategy

Use the formula T_n=a+\left(n-1\right)d.

Apply the idea

In the formula T_n=a+\left(n-1\right)d, \, d is the number multiplying (n-1). In our rule, T_n = 15+5\left(n-1\right), that number is 5.

d=5

c

Determine T_9, the 9th term in the sequence.

Worked Solution
Create a strategy

Substitute n=9 into the rule.

Apply the idea
\displaystyle T_9\displaystyle =\displaystyle 15+5\times\left(9-1\right)Substitute n=9
\displaystyle =\displaystyle 15+40Evaluate the multiplication
\displaystyle =\displaystyle 55Evaluate

Example 3

The first term of an arithmetic sequence is 2. The fifth term is 26.

a

Solve for d, the common difference of the sequence.

Worked Solution
Create a strategy

Substitute the values of a and T_5 into T_n=a+\left(n-1\right)d.

Apply the idea

We are given a=2 and T_5=26.

\displaystyle T_n\displaystyle =\displaystyle a+\left(n-1\right)dWrite the formula
\displaystyle T_5\displaystyle =\displaystyle a+(5-1)\times dSubstitute n=5
\displaystyle 26\displaystyle =\displaystyle 2+4 dSubstitute T_5=20, \, a=2 and simplify
\displaystyle 4d\displaystyle =\displaystyle 24Subtract 2 from both sides
\displaystyle d\displaystyle =\displaystyle 6Divide both sides by 4
b

Write a recursive rule for T_n in terms of T_{n-1} which defines this sequence and an initial condition for T_1.

Worked Solution
Create a strategy

Use the first term and common difference.

Apply the idea

We know that a=T_1=2. Since d=6 we know that each term is 6 more than the term before it.

So, we have:T_n=T_{n-1}+6,\,T_1=2

Example 4

In an arithmetic progression where a is the first term, and d is the common difference, T_7=44 andT_{14}=86.

a

Determine d, the common difference.

Worked Solution
Create a strategy

Find how many common differences need to be added to get from T_7 to T_{14}.

Apply the idea

There are 6 terms between T_7 and T_{14} with d being add each time: \begin{array}{c} &&+d&&+d&&+d&&+d&&+d&&+d&&+d \\ &T_7 & \longrightarrow &T_8 & \longrightarrow &T_9 & \longrightarrow &T_{10} & \longrightarrow &T_{11} & \longrightarrow &T_{12} & \longrightarrow& T_{13} & \longrightarrow &T_{14} \end{array}

So T_{14}=T_7+d+d+d+d+d+d=T_7+7d. We can solve this equation for d by substituting the values for T_7 and T_{14}.

\displaystyle T_{14}\displaystyle =\displaystyle T_7+7dWrite the equation
\displaystyle 86\displaystyle =\displaystyle 44+7dSubstitute the values
\displaystyle 7d\displaystyle =\displaystyle 42Subtract 44 from both sides
\displaystyle \dfrac{7d}{7}\displaystyle =\displaystyle \dfrac{42}{7}Divide both sides by 7
\displaystyle d\displaystyle =\displaystyle 6Evaluate
b

Determine a, the first term in the sequence.

Worked Solution
Create a strategy

Use the formula T_n=a+\left(n-1\right)d.

Apply the idea

We know that T_7=44, \, d=6, so we can substitute these values into the above formula and solve for a.

\displaystyle T_{7}\displaystyle =\displaystyle a+(7-1)\times dSubstitute n=7
\displaystyle 44\displaystyle =\displaystyle a+(7-1)\times 6Substitute T_7=44, \, d=6
\displaystyle 44\displaystyle =\displaystyle a+36Evaluate the multiplication
\displaystyle 44-36\displaystyle =\displaystyle a+36-36Subtract 36 from both sides
\displaystyle a\displaystyle =\displaystyle 8Evaluate
c

State the equation for T_n, the nth term in the sequence.

Worked Solution
Create a strategy

Substitute the values of a, and d, into the formula T_n=a+\left(n-1\right)d.

Apply the idea

We know that a=8 and d=6.

\displaystyle T_n\displaystyle =\displaystyle 8+(n-1)\times 6Substitute a=8, d=6
\displaystyle =\displaystyle 8+6n-6Expand the brackets
\displaystyle T_n\displaystyle =\displaystyle 6n+2Combine like terms
d

Hence find T_{25}, the 25th term in the sequence.

Worked Solution
Create a strategy

Substitute n=25 into the equation for T_n found from part (c).

Apply the idea
\displaystyle T_{25}\displaystyle =\displaystyle 6\times 25+2Substitute n=25
\displaystyle =\displaystyle 150+2Evaluate the multiplication
\displaystyle =\displaystyle 152Evaluate the addition
Idea summary

For any arithmetic sequence with starting value a and common difference d, we can express it in either of the following two forms:

  • Recursive form: t_n=t_{n-1}+d, where t_1=a

  • Explicit form: t_n=a+\left(n-1\right)d.

Arithmetic sequences in tables and graphs

For any arithmetic sequence in the general form given by t_n=a+\left(n-1\right)d, we can expand the bracket and collect like terms, creating a new generating rule of the form t_n=dn+k where d and k are constants. For example, the rule t_n=5+\left(n-1\right)\times 2 is equivalent to t_n=2n+3. This is in the form of the equation of a straight line \left(y=mx+c\right), so if an arithmetic sequence is plotted as a series of points, they will all lie on a straight line with the gradient being the common difference. This makes sense since we have a constant rate of change.

1
2
3
4
5
6
7
n
2
4
6
8
10
12
\text{Term Value}

The first term is represented by the point shown at n=1, \,t_1=5, and we can see the gradient here is the common difference d=2.

We could also be expected to recognise an arithmetic sequence from a table.

n12345
t_n5791113

We can read off the initial term t_1=5 and the common difference can be found from the difference between the t_n values.

Exploration

This interactive tool can show us how arithmetic sequences are actually linear relationships.

Loading interactive...

The common difference d determines the slope of the line. When the common difference is negative, the line has a negative slope. When the common difference is positive, the line has a positive slope.

As a increases, the y-intercept of the line increases.

Examples

Example 5

The nth term of an arithmetic progression is given by the equation T_n=12+4\left(n-1\right).

a

Complete the table of values.

n123410
T_n
Worked Solution
Create a strategy

Substitute the corresponding values of n into the equation.

Apply the idea
\displaystyle T_1\displaystyle =\displaystyle 12+4\times (1-1)Substitute n=1
\displaystyle =\displaystyle 12Evaluate
\displaystyle T_2\displaystyle =\displaystyle 12+4\times (2-1)Substitute n=2
\displaystyle =\displaystyle 16Evaluate
\displaystyle T_3\displaystyle =\displaystyle 12+4\times (3-1)Substitute n=3
\displaystyle =\displaystyle 20Evaluate
\displaystyle T_4\displaystyle =\displaystyle 12+4\times (4-1)Substitute n=4
\displaystyle =\displaystyle 24Evaluate
\displaystyle T_{10}\displaystyle =\displaystyle 12+4\times (10-1)Substitute n=10
\displaystyle =\displaystyle 48Evaluate

So, the complete table of values is given by:

n123410
T_n1216202448
b

By how much are consecutive terms in the sequence increasing?

Worked Solution
Create a strategy

Find the difference between the second term and the first term.

Apply the idea
\displaystyle \text{Common difference}\displaystyle =\displaystyle 16-12Subtract 12 from 16
\displaystyle =\displaystyle 4Evaluate
c

Plot the points in the table on the graph.

Worked Solution
Create a strategy

Plot the points from the table where n is on the horizontal axis, and T_n is on the vertical axis.

Apply the idea
2
4
6
8
10
12
n
10
20
30
40
T

From the table, we have the points: (1,12),\,(2,16),\,(3,20),\,(4,24),\,(10,48).

d

If the points on the graph were joined, they would form:

A
a straight line
B
a curve
Worked Solution
Create a strategy

Use the graph from part (c).

Apply the idea

The points on the graph lie in a straight line.

So, the correct answer is Option A.

Example 6

The plotted points represent terms in an arithmetic sequence:

1
2
3
4
5
n
2
4
6
8
10
12
14
T_n
a

Complete the table of values for the given points.

n1234
T_n
Worked Solution
Create a strategy

Use the coordinates of each point on the graph.

Apply the idea

The points on the graph have coordinates (1,1),\,(2,5),\,(3,9),\,(4,13). So the completed table is:

n1234
T_n15913
b

Identify d, the common difference between consecutive terms.

Worked Solution
Create a strategy

The common difference can be found by subtracting the first value from the second.

Apply the idea

We can see that the values of T_n increase by the same amount each time, so we can subtract any two consecutive terms to find the common difference.

\displaystyle d\displaystyle =\displaystyle 5-1Subtract T_1 from T_2
\displaystyle =\displaystyle 4Evaluate
c

Write a simplified expression for the general nth term of the sequence, T_n.

Worked Solution
Create a strategy

Substitute the values of a, the first term in the sequence, and d, the common difference, into the formula for finding the nth term: T_n=a+\left(n-1\right)d.

Apply the idea

We are given a=1 from the table and d=4 from part (b).

\displaystyle T_n\displaystyle =\displaystyle 1+(n-1)\times 4Substitute a=1, d=4
\displaystyle =\displaystyle 1+4n-4Expand the brackets
\displaystyle T_n\displaystyle =\displaystyle 4n-3Combine like terms
d

Find the 14th term of the sequence.

Worked Solution
Create a strategy

Substitute n=14 into the equation for T_n from part (c).

Apply the idea
\displaystyle T_{14}\displaystyle =\displaystyle 4\times 14-3Substitute n=14
\displaystyle =\displaystyle 53Evaluate
Idea summary

For any arithmetic sequence we can write a rule in the form \left(y=mx+c\right), so if an arithmetic sequence is plotted as a series of points, they will all lie on a straight line with the gradient being the common difference.

n is the independent variable and should lie on the horizontal axis. T_n is the dependent variable and should lie on the vertical axis.

Outcomes

ACMGM067

use recursion to generate an arithmetic sequence

ACMGM068

display the terms of an arithmetic sequence in both tabular and graphical form and demonstrate that arithmetic sequences can be used to model linear growth and decay in discrete situations

ACMGM069

deduce a rule for the nth term of a particular arithmetic sequence from the pattern of the terms in an arithmetic sequence, and use this rule to make predictions

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