9.02 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 $\left(u_{n+1}-u_n\right)$(un+1​−un​).

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 $u_1$u1​ and the common difference by the letter $d$d. Note that $t_n$tn​ or $T_n$Tn​ may also be used instead of $u_n$un​. 

We can find an explicit formula in terms of $u_1$u1​ 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 $u_1=-3$u1​=−3 and $d=8$d=8. A table of the sequence is show below:

$n$n	$u_n$un​	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 $u_{10}=69=-3+9\times8$u10​=69=−3+9×8  and the one-hundredth term $u_{100}=789=-3+99\times8$u100​=789=−3+99×8. And following the pattern, the explicit formula for the $n$nth term is $u_n=-3+(n-1)\times8$un​=−3+(n−1)×8.

We could create a similar table for the arithmetic progression with starting value $u_1$u1​ and common difference $d$d and we would observe the same pattern. Hence,  generating the explicit rule for any arithmetic sequence is given by:

 $u_n=u_1+\left(n-1\right)d$un​=u1​+(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 $u_1$u1​ and common difference $d$d and substitute these into the general form: $u_n=a+(n-1)d$un​=a+(n−1)d

Do: Each term is a decrease from the last by $7$7. So we have an arithmetic sequence with: $u_1=87$u1​=87 and $d=-7$d=−7. The general formula for this sequence is: $u_n=87+\left(n-1\right)\times\left(-7\right)$un​=87+(n−1)×(−7) or $u_n=87-7(n-1)$un​=87−7(n−1).

Hence, the $30$30th term is:

$u_n$un​	$=$=	$87-7\left(n-1\right)$87−7(n−1)	Write down the rule.
$\therefore\ u_{30}$∴ u30​	$=$=	$87-7\left(30-1\right)$87−7(30−1)	Substitute in $n=30$n=30.
	$=$=	$87-7\times29$87−7×29	Simplify.
	$=$=	$-116$−116	Evaluate.

 

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 $u_n$un​ and rearrange for $n$n. Finding $n$n is determining which term position has a value of $186$186.

Do: This is an arithmetic sequence with $u_1=10$u1​=10 and common difference $d=4$d=4. Hence, the general rule is: $u_n=10+\left(n-1\right)\times4$un​=10+(n−1)×4, we can simplify this to $u_n=6+4n$un​=6+4n, by expanding brackets and collecting like terms. Substituting $u_n=186$un​=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.

 

Practice questions

Question 1

Question 2

Question 3