VCE General Mathematics 1&2 - 2020 Edition
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6.03 Arithmetic sequences
Lesson

What is an 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 $t_{n+1}-t_n$tn+1tn.

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

The first term in an arithmetic sequence is denoted by the letter $a$a and the common difference is denoted by $d$d. Since, $t_2=t_1+d$t2=t1+d$t_3=t_2+d$t3=t2+d and so on, any arithmetic sequence can be expressed as the recurrence relation that was defined in the last lesson of this chapter:

$t_{n+1}=t_n+d,t_1=a$tn+1=tn+d,t1=a

An explicit generating rule can be found in terms of $a$a and $d$d, this is useful for finding the $n$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$3,5,13,21,, the starting term is $-3$3 and there is a common difference of $8$8, that is $a=-3$a=3 and $d=8$d=8. A table of the sequence is show below:

$n$n $t_n$tn Pattern
$1$1 $-3$3 $-3$3
$2$2 $5$5 $-3+8$3+8
$3$3 $13$13 $-3+2\times8$3+2×8
$4$4 $21$21 $-3+3\times8$3+3×8
...    
$n$n $t_n$tn $-3+(n-1)\times8$3+(n1)×8

By correctly identifying the pattern, the tenth term becomes $t_{10}=69=-3+9\times8$t10=69=3+9×8  and the one-hundredth term would be $t_{100}=789=-3+99\times8$t100=789=3+99×8. Following the pattern, the explicit formula for the $n$nth term is $t_n=-3+(n-1)\times8$tn=3+(n1)×8.

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

 $t_n=a+\left(n-1\right)d$tn=a+(n1)d

Forms of arithmetic sequences

For any arithmetic sequence with starting value $a$a and common difference $d$d, 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:

$t_{n+1}=t_n+d$tn+1=tn+d, where $t_1=a$t1=a

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

$t_n=a+\left(n-1\right)d$tn=a+(n1)d

Worked examples

Example 1

For the sequence  $87,80,73,66...$87,80,73,66..., find and explicit rule for the $n$nth term and hence, find the $30$30th term. 

Think: Check that the sequence is arithmetic, does each term differ from the last by a constant? Then write down the the starting value $a$a and common difference $d$d and substitute these into the general form: $t_n=a+(n-1)d$tn=a+(n1)d

Do: Each term is a decrease from the last by $7$7. So we have an arithmetic sequence with: $a=87$a=87 and $d=-7$d=7. The general formula for this sequence is: $t_n=87+\left(n-1\right)\times\left(-7\right)$tn=87+(n1)×(7) or $t_n=87-7(n-1)$tn=877(n1).

Hence, the $30$30th term is: $t_{30}=87-7\times29=-116$t30=877×29=116.

Example 2

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

Think: Find a general rule for the sequence, substitute in $186$186 for $t_n$tn and rearrange for $n$n.

Do: This is an arithmetic sequence with $a=10$a=10 and common difference $d=4$d=4. Hence, the general rule is: $t_n=10+\left(n-1\right)\times4$tn=10+(n1)×4, we can simplify this to $t_n=6+4n$tn=6+4n, by expanding brackets and collecting like terms. Substituting $t_n=186$tn=186, we get:

$186$186 $=$= $6+4n$6+4n
$\therefore4n$4n $=$= $180$180
$n$n $=$= $45$45

Hence, the $45$45th term in the sequence is $186$186.

Example 3

If an arithmetic sequence has $t_5=38$t5=38 and $t_9=66$t9=66, find the recurrence relation for the sequence. 

Think: To find the recurrence relation we need the starting value and common difference. As we have two terms we can set up two equations in terms of $a$a and $d$d using $t_n=a+(n-1)d$tn=a+(n1)d.

Do:

$t_5$t5: $a+4d=38$a+4d=38 $.....\left(1\right)$.....(1)

and

$t_9$t9: $a+8d=66$a+8d=66 $.....\left(2\right)$.....(2)

 

If we now subtract equation $\left(1\right)$(1) from equation $\left(2\right)$(2) the first term in each equation will cancel out to leave us with:  

$\left(8d-4d\right)$(8d4d) $=$= $66-38$6638
$4d$4d $=$= $28$28
$\therefore d$d $=$= $7$7

With the common difference found to be $7$7, then we know that, using equation $\left(1\right)$(1) $a+4\times7=38$a+4×7=38 and so $a$a is $10$10. The recurrence relation for this sequence is given by:  

$t_{n+1}=t_n+7,t_1=10$tn+1=tn+7,t1=10

Practice questions

QUESTION 1

The $n$nth term in an arithmetic progression is given by the formula $T_n=15+5\left(n-1\right)$Tn=15+5(n1).

  1. Determine $a$a, the first term in the arithmetic progression.

  2. Determine $d$d, the common difference.

  3. Determine $T_9$T9, the $9$9th term in the sequence.

QUESTION 2

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

  1. Solve for $d$d, the common difference of the sequence.

  2. Write a recursive rule for $T_n$Tn in terms of $T_{n-1}$Tn1 which defines this sequence and an initial condition for $T_1$T1.

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

QUESTION 3

In an arithmetic progression where $a$a is the first term, and $d$d is the common difference, $T_7=43$T7=43 and $T_{14}=85$T14=85.

  1. Determine $d$d, the common difference.

  2. Determine $a$a, the first term in the sequence.

  3. State the equation for $T_n$Tn, the $n$nth term in the sequence.

  4. Hence find $T_{25}$T25, the $25$25th term in the sequence.

 

 

Arithmetic sequences in tables and graphs

For any arithmetic sequence in the general form given by $t_n=a+\left(n-1\right)d$tn=a+(n1)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 $t_n=dn+k$tn=dn+k where $d$d and $k$k are constants. For example, the rule $t_n=5+\left(n-1\right)\times2$tn=5+(n1)×2 is equivalent to $t_n=2n+3$tn=2n+3. This is in the form of the equation of a straight line $\left(y=mx+c\right)$(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 gradient being the common difference. This makes sense as there is a constant rate of change, i.e. the common difference.

 

The first term is represented by the point shown at $n=1$n=1$t_1=5$t1=5 and the gradient of this line is the common difference $d=2$d=2

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

n 1 2 3 4 5
tn 5 7 9 11 13

Here, the initial term $t_1=5$t1=5 and the common difference can be seen in step between the $t_n$tn values in the second row. Since $2$2 is being added each time to create the next term in the sequence, the common difference is $d=2$d=2.

 

Practice questions

QUESTION 4

The $n$nth term of an arithmetic progression is given by the equation $T_n=12+4\left(n-1\right)$Tn=12+4(n1).

  1. Complete the table of values.

    $n$n $1$1 $2$2 $3$3 $4$4 $10$10
    $T_n$Tn $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. By how much are consecutive terms in the sequence increasing?

  3. Plot the points in the table on the graph.

    Loading Graph...

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

    a straight line

    A

    a curved line

    B

    a straight line

    A

    a curved line

    B

QUESTION 5

The plotted points represent terms in an arithmetic sequence:

Loading Graph...

  1. Complete the table of values for the given points.

    $n$n $1$1 $2$2 $3$3 $4$4
    $T_n$Tn $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. Identify $d$d, the common difference between consecutive terms.

  3. Write a simplified expression for the general $n$nth term of the sequence, $T_n$Tn.

  4. Find the $15$15th term of the sequence.

QUESTION 6

The given table of values represents terms in an arithmetic sequence.

$n$n $1$1 $2$2 $3$3 $4$4
$T_n$Tn $9$9 $17$17 $25$25 $33$33
  1. Identify $d$d, the common difference between consecutive terms.

  2. Write a simplified expression for the general $n$nth term of the sequence, $T_n$Tn.

  3. Find the $15$15th term of the sequence.

Outcomes

AoS3.12

Define and explain key concepts in use of a first-order linear recurrence relation to generate the terms of a number sequence, and apply a range of related mathematical routines and procedures

AoS3.14

Define and explain key concepts in generation of an arithmetic sequence using a recurrence relation, tabular and graphical display; and the rule for the nth term of an arithmetic sequence and its evaluation, and apply a range of related mathematical routines and procedures

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