3.03 Arithmetic progressions - calculator free

 

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 $\left(t_{n+1}-t_n\right)$(tn+1​−tn​).

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

We denote the first term by the letter $a$a and the common difference by the letter $d$d. Since, $t_2=t_1+d$t2​=t1​+d, $t_3=t_2+d$t3​=t2​+d and so on, we can write any arithmetic sequence as the recurrence relation:

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

If we consider the recurrence relation $t_n=t_{n-1}+2,t_1=5$tn​=tn−1​+2,t1​=5, this recurrence relation starts with $5$5 and we add $2$2 to find the next term, therefore the sequence is $5,7,9,11,13,15...$5,7,9,11,13,15...

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

Let's look at a table of a concrete example to see the pattern for the explicit formula. For the sequence $-3,5,13,21,\ldots$−3,5,13,21,…, we have starting term of $-3$−3 and 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+(n−1)×8

 

 

The pattern starts to become clear and we could guess that the tenth term becomes $t_{10}=69=-3+9\times8$t10​=69=−3+9×8  and the one-hundredth term $t_{100}=789=-3+99\times8$t100​=789=−3+99×8. And following the pattern, the explicit formula for the $n$nth term is $t_n=-3+(n-1)\times8$tn​=−3+(n−1)×8.

We could create a similar table for the arithmetic progression with starting value $a$a and common difference $d$d and we would observe the same pattern. Hence,  generating the explicit rule for any arithmetic sequence is given by:

 $t_n=a+\left(n-1\right)d$tn​=a+(n−1)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+(n−1)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+(n−1)×(−7) or $t_n=87-7(n-1)$tn​=87−7(n−1).

Hence, the $30$30th term is:$t_n=87+\left(30-1\right)\times\left(-7\right)$tn​=87+(30−1)×(−7)$t_{30}=87-7\times29=-116$t30​=87−7×29=−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. Finding $n$nis determining which term position has a value of 186.

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+(n−1)×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: Consider how many times the common difference would need to be added to get from $T_5$T5​ to $T_9$T9​.

$T_5$T5​		$T_6$T6​		$T_7$T7​		$T_8$T8​		$T_9$T9​
	$+d$+d		$+d$+d		$+d$+d		$+d$+d

 

Do: The common difference needs to be added to $T_5$T5​ four times to get to $T_9$T9​. Therefore $T_5+4d=T_9$T5​+4d=T9​ so $38+4d=66$38+4d=66, this can be rearranged to solve for $d$d. Alternatively, we can use logic, an increase from $38$38 to $66$66 is made up of $4$4 common differences, so $d=(66-38)/4$d=(66−38)/4. 

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

$T_n=T_{n-1}+7,T_1=10$Tn​=Tn−1​+7,T1​=10

 

Practice questions

Question 1

Question 2

Question 3

 

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+(n−1)d, we can expand the bracket and collect like terms, 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+(n−1)×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, 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.

The first term is represented by the point shown at $n=1$n=1, $t_1=5$t1​=5 and we can see the gradient here is the common difference $d=2$d=2. 

We could also be expected to recognise an arithmetic sequence from a table, such as:

$n$n	$1$1	$2$2	$3$3	$4$4	$5$5
$t_n$tn​	$5$5	$7$7	$9$9	$11$11	$13$13

Here, we can read off 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.

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

Created with Geogebra

 

Practice questions

Question 4

Question 5

Question 6