8.04 Arithmetic series (Enrichment)

The definition of a series

The sum of terms in a sequence is called a series. The sum of the first $n$n terms of a sequence is denoted $S_n$Sn​, which can be written as $S_n=u_1+u_2+u_3+\ldots+u_n$Sn​=u1​+u2​+u3​+…+un​. We can find a specific term in an arithmetic series just as we would in a sequence using the formula $u_n=u_1+(n-1)d$un​=u1​+(n−1)d.

Summing $n$n terms in an arithmetic sequence

Adding a significant number of terms without a shortcut would be very tedious, so let's develop a formula for calculating the total of an arithmetic series.

What would be the sum of the first $100$100 integers? An oft quoted anecdote has the famous mathematician Carl Friedrich Gauss, at the age of seven, who was set this task by his teacher to add up the first $100$100 integers. The teacher is said to have set the task to keep Gauss busy for a while, but he replied almost immediately with the answer. Let's look at the way this was calculated.

First, we write down the sum in order:

$S_{100}=1+2+3+4+5+\ldots+96+97+98+99+100$S100​=1+2+3+4+5+…+96+97+98+99+100

Then underneath we write the sum in reverse order:

$S_{100}=$S100​=	$1$1	$+2$+2	$+3$+3	$+4$+4	$...$...	$+97$+97	$+98$+98	$+99$+99	$+100$+100
$S_{100}=$S100​=	$100$100	$+99$+99	$+98$+98	$+97$+97	$...$...	$+4$+4	$+3$+3	$+2$+2	$+1$+1

Adding term-wise vertically we get the sum:

$2S_{100}=101+101+101+101+\ldots+101+101+101+101$2S100​=101+101+101+101+…+101+101+101+101

Since we added $100$100 terms we have:

$2S_{100}=100\times101$2S100​=100×101

And hence:

$S_{100}=\frac{100\times101}{2}=5050$S100​=100×1012​=5050.

So the first $100$100 integers add to $5050$5050. This was an arithmetic sequence with $u_1=1$u1​=1 and $d=1$d=1. But we could mimic this process for a general arithmetic sequence of length $n$n with starting value $u_1$u1​ and common difference $d$d.

$S_n=u_1+\left(u_1+d\right)+\left(u_1+2d\right)+....+\left(u_1+\left(n-3\right)d\right)+\left(u_1+\left(n-2\right)d\right)+\left(u_1+\left(n-1\right)d\right)$Sn​=u1​+(u1​+d)+(u1​+2d)+....+(u1​+(n−3)d)+(u1​+(n−2)d)+(u1​+(n−1)d)

 

Then writing this sum again underneath but in reverse order, we get:

$S_n=u_1$Sn​=u1​	$+\left(u_1+d\right)$+(u1​+d)	$+\left(u_1+2d\right)$+(u1​+2d)	$+....$+....	$+\left(u_1+\left(n-3\right)d\right)$+(u1​+(n−3)d)	$+\left(u_1+\left(n-2\right)d\right)$+(u1​+(n−2)d)	$+\left(u_1+\left(n-1\right)d\right)$+(u1​+(n−1)d)
$S_n=\left(u_1+\left(n-1\right)d\right)$Sn​=(u1​+(n−1)d)	$+\left(u_1+\left(n-2\right)d\right)$+(u1​+(n−2)d)	$+\left(u_1+\left(n-3\right)d\right)$+(u1​+(n−3)d)	$+....$+....	$+\left(u_1+2d\right)$+(u1​+2d)	$+\left(u_1+d\right)$+(u1​+d)	$+a$+a

 

Adding term-wise vertically each pair of terms adds to $2u_1+\left(n-1\right)d$2u1​+(n−1)d and we have $n$n of these terms so:

$2S_n=n\left(2u_1+\left(n-1\right)d\right)$2Sn​=n(2u1​+(n−1)d)

And hence, the sum of the first $n$n terms of an arithmetic sequence is given by:

 $S_n=\frac{n}{2}\left(2+\left(n-1\right)d\right)$Sn​=n2​(2+(n−1)d)

Once we know the first term $u_1$u1​, the common difference $d$d and the number of terms $n$n that we wish to add, we are able to calculate the value of an arithmetic series $S_n$Sn​.

If we know the last term of the arithmetic sequence, then the task becomes even easier. Splitting the $u_1$u1​'s up in the formula we can write $S_n$Sn​ as:

$S_n=\frac{n}{2}\left[u_1+u_1+\left(n-1\right)d\right]$Sn​=n2​[u1​+u1​+(n−1)d]

In any series $S_n$Sn​, the last term will be $u_n$un​, which we know is:

$l=u_n=u_1+\left(n-1\right)d$l=un​=u1​+(n−1)d

And so we can deduce that:

$S_n=\frac{n}{2}\left[u_1+l\right]$Sn​=n2​[u1​+l]

This formula is much simpler to work with, without the 'brackets inside the brackets' present in the first form. 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

Applications of arithmetic series

Questions may involve algebraic manipulation and simultaneous equations to solve for $u_1$u1​, $n$n or $d$d before we are able to calculate our sum. When solving for $n$n, take care because $n$n appears twice in the formula $S_n=\frac{n}{2}\left(2u_1+\left(n-1\right)d\right)$Sn​=n2​(2u1​+(n−1)d), meaning we will need to solve a quadratic equation. Although this will result in two solutions, often, one is a negative value of $n$n. This isn't possible, seeing as $n$n must be a positive integer.

In application questions, there may be a mixture of parts requiring the use of both the $u_n$un​ and $S_n$Sn​ formulae. Think carefully, "is this question asking me about a specific term of a sequence, or the total of many terms of the sequence?" Let's consider some worked examples and practice questions. 

Worked examples

Example 1

Rows of stands to take school photos are set up with $2$2 additional places in each successive row.

(a) If there are $10$10 students in the first row how many students in total would there be in $5$5 rows?

Think: We want to sum of the first $5$5 terms of an arithmetic sequence that starts with $10$10 and the common difference is $2$2. Write down $n$n, $u_1$u1​ and $d$d and use the formula $S_n=\frac{n}{2}\left(2u_1+\left(n-1\right)d\right)$Sn​=n2​(2u1​+(n−1)d) to find the sum.

Do: $n=5$n=5, $u_1=10$u1​=10 and $d=2$d=2, substituting into the formula we get:

$S_5$S5​	$=$=	$\frac{5}{2}\left(2\times10+\left(5-1\right)2\right)$52​(2×10+(5−1)2)
	$=$=	$\frac{5}{2}\left(20+8\right)$52​(20+8)
	$=$=	$70$70

There would be $70$70 students in the photo.

(b) If there were $15$15 students in the first row, how many rows are required for a shot of a year level with $159$159 students?

Think: We have the sum but require $n$n. We need to write down $u_1$u1​, $d$d, and $S_n$Sn​, substitute these values into $S_n=\frac{n}{2}\left(2u_1+\left(n-1\right)d\right)$Sn​=n2​(2u1​+(n−1)d), and solve for $n$n. Note: Since $n$n appears in two factors of this formula we will obtain a quadratic equation. We can then solve this by factorisation, the quadratic formula or an appropriate method using technology.

Do: $u_1=15,d=2,S_n=147$u1​=15,d=2,Sn​=147 and hence:

$S_n$Sn​	$=$=	$\frac{n}{2}\left(2u_1+\left(n-1\right)d\right)$n2​(2u1​+(n−1)d)
$147$147	$=$=	$\frac{n}{2}\left(2\times15+\left(n-1\right)2\right)$n2​(2×15+(n−1)2)	Substitute in values
$294$294	$=$=	$n\left(28+2n\right)$n(28+2n)	Multiply both sides by $2$2 and simplify
$294$294	$=$=	$28n+2n^2$28n+2n2	Expand brackets
$\therefore2n^2+28n-294$∴2n2+28n−294	$=$=	$0$0	Bring all terms to one side
$n^2+14n-147$n2+14n−147	$=$=	$0$0	Dividing both sides by $2$2
$(n-7)(n+21)$(n−7)(n+21)	$=$=	$0$0	Factorise

 

Hence, $n=7$n=7 as $n>0$n>0. So they will require $7$7 rows for the photo.

Example 2

Find the sum of the multiples of $7$7 between $100$100 and $200$200.

Think: What is the first multiple of seven greater than $100$100, what is the last multiple of seven before $200$200 and how many multiples are there?

Do: By dividing the numbers $101,102,103,...$101,102,103,... by seven we find the first multiple above $100$100 is $u_1=105$u1​=105 and similarly the last multiple before $200$200 is $l=196$l=196. The number of multiples, $n$n, can be found from dividing the difference between our first and last value by $7$7 and adding $1$1, to count our first multiple. Hence, $n=\frac{196-105}{7}+1=14$n=196−1057​+1=14.

Alternatively, this can be found by considering we have the sequence $105,112,119,...,196$105,112,119,...,196, with $u_1=105$u1​=105, $d=7$d=7 and solving $u_n=196$un​=196 for $n$n using $u_n=u_1+(n-1)d$un​=u1​+(n−1)d. This gives us

$196$196	$=$=	$105+\left(n-1\right)\times7$105+(n−1)×7
	$=$=	$105+7n-7$105+7n−7
	$=$=	$98+7n$98+7n
$98$98	$=$=	$7n$7n
$n$n	$=$=	$14$14

 

For the summation, using $S_n=\frac{n}{2}\left(u_1+l\right)$Sn​=n2​(u1​+l), we have:

$S_{14}$S14​	$=$=	$\frac{14}{2}\left(105+196\right)$142​(105+196)
	$=$=	$2100$2100

Practice questions

QUESTION 4

QUESTION 5