6.02 Introduction to arithmetic progressions

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. So, $u_2=u_1+d$u2​=u1​+d, $u_3=u_2+d$u3​=u2​+d and so on. We can also 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	$u_n$un​	$-3+(n-1)\times8$−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, the generating 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: $u_n=u_1+(n-1)d$un​=u1​+(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: $u_{30}=87-7\times29=-116$u30​=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 $u_n$un​ and rearrange for $n$n.

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.

Example 3

If an arithmetic sequence has $u_5=38$u5​=38 and $u_9=66$u9​=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 $u_1$u1​ and $d$d using $u_n=u_1+(n-1)d$un​=u1​+(n−1)d.

Do:

$u_5$u5​:

$u_1+4d=38$u1​+4d=38

$.....\left(1\right)$.....(1)

and

$u_9$u9​:

$u_1+8d=66$u1​+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:  

$\left(8d-4d\right)$(8d−4d)	$=$=	$66-38$66−38
$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) $u_1+4\times7=38$u1​+4×7=38 and so $u_1$u1​ is $10$10.

Practice questions

QUESTION 1

 

QUESTION 2

 

Applications of arithmetic sequences

We have seen many applications of linear growth or decay when studying linear functions in Chapter 2. Hence, arithmetic sequences apply in many areas of life, including simple interest earnings, straight-line depreciation, monthly rental accumulation and many others. For example, if you are saving money in equal instalments, the cumulative savings at each savings period form an arithmetic sequence. If you are travelling down a highway at a constant speed, the amount of petrol left in the tank, if measured every minute of the trip, forms another arithmetic progression. In fact any time you notice a quantity changing in equal amounts at set time periods, then you can consider that process as being arithmetic.

Worked example

Example 4

Tabitha starts with $\$200$$200 in her piggy bank, the following week she adds $\$25$$25 and then continues to add $\$25$$25 at the start of each successive week. Find a rule to describe $B_n$Bn​ the balance of her savings at the start of each week and find when her savings will reach $\$450$$450.

Think: The sequence of savings generated is $\$200,\$225,\$250,\$275...$$200,$225,$250,$275...  This is arithmetic, so write down the starting value a and common difference and use $u_n=u_1+\left(n-1\right)d$un​=u1​+(n−1)d to find a general rule.

Do: $u_1=200$u1​=200 and $d=25$d=25 and so our general rule is: $B_n=200+25(n-1)$Bn​=200+25(n−1) or equivalently $B_n=175+25n$Bn​=175+25n.

To find when the savings reach $\$450$$450, we substitute this into our general rule and solve for $n$n:

$450$450	$=$=	$175+25n$175+25n
$\therefore25n$∴25n	$=$=	$275$275
$n$n	$=$=	$11$11

Hence, at the start of the $11$11th week her savings will have grown to $\$450$$450.

Practice questions

QUESTION 3