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5. Number Patterns and Recursions
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VCE 11 General 2023

5.03 Arithmetic sequences

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Arithmetic sequence

A sequence in which each term changes from the last by adding or subtracting a constant amount is called an arithmetic sequence. The number being added or subtracted to produce the next number in the sequence is known as the common difference, which will result from subtracting any two successive terms u_{n + 1}- u_n.

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

The first term in an arithmetic sequence is denoted by the letter a and the common difference is denoted by d. Since, u_1=u_0+d, u_2=u_1+d, and so on, any arithmetic sequence can be expressed as the recurrence relation that was defined in the  last lesson of this chapter  :

u_{n+1}=u_n + d, u_0=a

An explicit generating rule can be found in terms of a and d, this is useful for finding the nth term without listing the sequence or having to use the previous term in the sequence each time to find the next term.

Consider the following table to see the pattern for the explicit formula. For the sequence -3,5,13,21, \ldots, the starting term is -3 and there is a common difference of 8, that is a=-3 and d=8. A table of the sequence is show below:

nu_n\text{Pattern}
1-3-3
25-3+8
313-3+2\times8
421-3+3\times8
\ldots
nu_{n+1}-3+(n-1)\times8

By correctly identifying the pattern, the tenth term becomes u_{10}=69=-3+9\times 8 and the one-hundredth term would be u_{100}=789=-3+99\times 8. Following the pattern, the explicit formula for the nth term is u_n=-3+(n-1)\times 8.

A similar table can be created for any arithmetic sequence with starting value a and common difference d and the same pattern would be observed. Hence, the explicit generating rule for the nth term in any arithmetic sequence is given by:

u_n=a+(n-1)\times d

Examples

Example 1

Consider the following sequence.

87, \,80, \,73, \,66,...

a

Find an explicit rule for the nth term.

Worked Solution
Create a strategy

Find the common difference, d, to substitute into u_n=a+(n-1)\times d.

Apply the idea

Since 87-7=80, 80-7=73, and so on, d=-7.

The first term is 87, so a=87.

\displaystyle u_n\displaystyle =\displaystyle a+(n-1)\times dWrite the formula
\displaystyle u_n\displaystyle =\displaystyle 87+(n-1)\times -7Substitute a=87 and d=-7
b

Hence, find the 30th term.

Worked Solution
Create a strategy

Substitute n=30 into u_n=87+(n-1)\times -7 .

Apply the idea
\displaystyle u_n\displaystyle =\displaystyle 87+(n-1)\times -7Write the formula
\displaystyle u_{30}\displaystyle =\displaystyle 87+(30-1)\times -7Substitute n=30
\displaystyle =\displaystyle 87+29\times -7Evaluate the subtraction
\displaystyle =\displaystyle 87-203Evaluate the multiplication
\displaystyle =\displaystyle -116Evaluate

Example 2

For the sequence 10, 14, 18, 22, 26,..., find n if the nth term is 186.

Worked Solution
Create a strategy

Substitute T_n=186, a=10, and value of d into T_n=a+(n-1)\times d.

Apply the idea

Since 10+4=14, 14+4=18, and so on, d=4.

\displaystyle T_n\displaystyle =\displaystyle a+(n-1)\times dWrite the formula
\displaystyle 186\displaystyle =\displaystyle 10+(n-1)\times 4Substitute T_n=186,a=10, and d=4
\displaystyle 186\displaystyle =\displaystyle 10+4n-4Expand the brackets
\displaystyle 186\displaystyle =\displaystyle 6+4nCollect like terms
\displaystyle 186-6\displaystyle =\displaystyle 6+4n-6Subtract 6 from both sides
\displaystyle 180\displaystyle =\displaystyle 4nEvaluate
\displaystyle \dfrac{180}{4}\displaystyle =\displaystyle \dfrac{4n}{4}Divide both sides by 4
\displaystyle 45\displaystyle =\displaystyle nEvaluate
\displaystyle n\displaystyle =\displaystyle 45Make n the subject

Example 3

The nth term in an arithmetic progression is given by the formula T_n=15+5\times\left(n-1\right).

a

Determine a, the first term in the arithmetic progression.

Worked Solution
Create a strategy

Since a is the first term or when n=1, substitute T_n=a and n=1 into T_n=15+5\times\left(n-1\right).

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle 15+5\times\left(n-1\right)Write the formula
\displaystyle a\displaystyle =\displaystyle 15+5\times\left(1-1\right)Substitute T_n=a and n=1
\displaystyle a\displaystyle =\displaystyle 15+5\times(0)Evaluate the subtraction
\displaystyle a\displaystyle =\displaystyle 15+0Evaluate the multiplication
\displaystyle a\displaystyle =\displaystyle 15Evaluate
b

Determine d, the common difference.

Worked Solution
Create a strategy

Compare the explicit formula u_n=a+(n-1)\times d to T_n=15+5\times\left(n-1\right).

Apply the idea

The explicit formula shows that d is the number multiplied by (n-1).

So d=5.

c

Determine T_9, the 9th term in the sequence.

Worked Solution
Create a strategy

Substitute n=9 into T_n=15+5\times\left(n-1\right).

Apply the idea
\displaystyle T_9\displaystyle =\displaystyle 15+5\times(9-1)Substitute n=9
\displaystyle =\displaystyle 15+5\times(8)Evaluate the subtraction
\displaystyle =\displaystyle 15+40Evaluate the multiplication
\displaystyle =\displaystyle 55Evaluate

Example 4

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

Construct an equation between 26 and the steps to get into it.

Apply the idea

To get into the fifth term, 26, we need to add 2 to 4 times the common difference.

2+4d=26

Use the equation to solve for d.

\displaystyle 2+4d\displaystyle =\displaystyle 26Write the formula
\displaystyle 2+4d-2\displaystyle =\displaystyle 26-2Subtract 2 from both sides
\displaystyle 4d\displaystyle =\displaystyle 24Evaluate
\displaystyle \dfrac{4d}{4}\displaystyle =\displaystyle \dfrac{24}{4}Divide both sides by 4
\displaystyle d\displaystyle =\displaystyle 6Evaluate
b

Write a recursive rule for u_{n+1} in terms of u_n which defines this sequence and an initial condition for u_0.

Write both parts on the same line separated by a comma.

Worked Solution
Create a strategy

Substitute d=6 and a=2 into the recursive formula u_{n+1}=u_n+d,u_0=a.

Apply the idea
\displaystyle u_{n+1}\displaystyle =\displaystyle u_n+d,u_0=aWrite the formula
\displaystyle u_{n+1}\displaystyle =\displaystyle u_n+6,u_0=2Substitute d=6 and a=2

Example 5

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

a

Determine d, the common difference.

Worked Solution
Create a strategy

Divide the difference of T_7 from T_{14} by the number of steps between them.

Apply the idea

The difference of two terms is 86-44=42.

The number of steps between T_7 and T_{14} is 14-7=7.

\displaystyle d\displaystyle =\displaystyle \dfrac{42}{7}Divide the difference by the number of steps
\displaystyle =\displaystyle 6Evaluate
b

Determine a, the first term in the sequence.

Worked Solution
Create a strategy

Find the product of d and number of steps between T_1 and T_7 to subtract from T_7=44.

Apply the idea

The number of steps from T_1 to T_7 is 6, so multiplying it by d=6 will give 6\times6.

\displaystyle a\displaystyle =\displaystyle 44-36Subtract the product from T_7=44
\displaystyle =\displaystyle 8Evaluate
c

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

Worked Solution
Create a strategy

Substitute a=8 and d=6 into the explicit formula T_n=a+(n-1)\times d.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle a+(n-1)\times dWrite the formula
\displaystyle T_n\displaystyle =\displaystyle 8+6\times(n-1)Substitute a=8 and d=6
\displaystyle T_n\displaystyle =\displaystyle 8+6n-6Expand the brackets
\displaystyle T_n\displaystyle =\displaystyle 2+6nCollect like terms
d

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

Worked Solution
Create a strategy

Substitute n=25 into T_n=6n+2.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle 2+6nWrite the formula
\displaystyle T_{25}\displaystyle =\displaystyle 2+6\times25Substitute n=25
\displaystyle T_{25}\displaystyle =\displaystyle 2+150Evaluate the multiplication
\displaystyle T_{25}\displaystyle =\displaystyle 152Evaluate
Idea summary

For any arithmetic sequence with starting valuea and common differenced, the sequence can be expressed in either of the following two forms:

  • Recursive form is a way to express any term in relation to the previous term:

    \displaystyle u_{n+1} = u_{n} + d,u_0 = a
    \bm{d}
    is the common difference
    \bm{a}
    is the first term

  • Explicit form is a way to express any term in relation to the term number:

    \displaystyle u_n = a + (n-1)\times d
    \bm{a}
    is the first term
    \bm{d}
    is the common difference

Arithmetic sequences in tables and graphs

For any arithmetic sequence in the general form given by u_n=a+(n-1)\times d, the right-hand side of the equation can be expanded using the distributive law and then like terms can be collected, creating a new generating rule of the form u_n=dn+k where d and k are constants.

For example, the rule u_n=5+(n-1)\times 2 is equivalent to u_n=2n+3. This is in the form of the equation of a straight line y=mx+c, so if an arithmetic sequence is plotted as a series of points, all the points lie on a straight line with the slope being the common difference. This makes sense as there is a constant rate of change, i.e. the common difference.

1
2
3
4
5
n
2
4
6
8
10
12
14
u_n

The first term is represented by the point shown at n=1,u_1=5, and the slope of this line is the common difference d=2.

An arithmetic sequence can be identified from a table, such as:

n12345
u_n5791113

Here, the initial term u_1=5 and the common difference can be seen in step between the u_n values in the second row. Since 2 is being added each time to create the next term in the sequence, the common difference is d=2.

Examples

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

Put the corresponding T_n-value of each n-value from the graph.

Apply the idea
n1234
T_n15913
b

Identify d, the common difference between consecutive terms.

Worked Solution
Create a strategy

Subtract the value of T_1 from T_2.

Apply the idea
\displaystyle d\displaystyle =\displaystyle 5-1Subtract T_1=1 from T_2=5
\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 first term, a=1, and d=4 into the explicit formula T_n=a+(n-1)\times d.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle a+(n-1)\times dWrite the formula
\displaystyle T_n\displaystyle =\displaystyle 1+(n-1)\times 4Substitute a=1 and d=4
\displaystyle T_n\displaystyle =\displaystyle 1+4n-4Expand the brackets
\displaystyle T_n\displaystyle =\displaystyle -3+4nCollect like terms
d

Find the 14th term of the sequence.

Worked Solution
Create a strategy

Substitute n=14 into T_n=-3+4n.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle -3+4nWrite the formula
\displaystyle T_{14}\displaystyle =\displaystyle -3+4\times14Substitute n=14
\displaystyle T_{14}\displaystyle =\displaystyle -3+56Evaluate the multiplication
\displaystyle T_{14}\displaystyle =\displaystyle 53Evaluate
Idea summary

The following are the steps of evaluating an arithmetic sequence from a graph.

  1. Make a table of values of the corresponding n and T_n-values.

  2. Identify the common difference, d, by subtracting the first term from the second term.

  3. Write the general equation substituting a=T_1 and d into the explicit formula T_n=a+(n-1)\times d.

  4. Use the formula to find the required nth term.

Outcomes

U1.AoS2.7

use a given recurrence relation to generate a sequence, deduce the explicit rule, n u from the recursion relation, tabulate, graph and evaluate the sequence

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