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 $S_n=T_1+T_2+T_3+\ldots+T_n$Sn=T1+T2+T3+…+Tn. We can find a specific term in an arithmetic series just as we would in a sequence using the formula $T_n=a+(n-1)d$Tn=a+(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 $a=1$a=1 and $d=1$d=1. But we could mimic this process for a general arithmetic sequence of length $n$n with starting value $a$a and common difference $d$d.
$S_n=a+\left(a+d\right)+\left(a+2d\right)+....+\left(a+\left(n-3\right)d\right)+\left(a+\left(n-2\right)d\right)+\left(a+\left(n-1\right)d\right)$Sn=a+(a+d)+(a+2d)+....+(a+(n−3)d)+(a+(n−2)d)+(a+(n−1)d)
Then writing this sum again underneath but in reverse order, we get:
Adding term-wise vertically each pair of terms adds to $2a+\left(n-1\right)d$2a+(n−1)d and we have $n$n of these terms so:
$2S_n=n\left(2a+\left(n-1\right)d\right)$2Sn=n(2a+(n−1)d)
And hence, the sum of the first $n$n terms of an arithmetic sequence is given by:
$S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$Sn=n2(2a+(n−1)d)
Once we know the first term $a$a, 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 $a$a's up in the formula we can write $S_n$Sn as:
$S_n=\frac{n}{2}\left[a+a+\left(n-1\right)d\right]$Sn=n2[a+a+(n−1)d]
In any series $S_n$Sn, the last term will be $T_n$Tn, which we know is:
$l=T_n=a+\left(n-1\right)d$l=Tn=a+(n−1)d
And so we can deduce that:
$S_n=\frac{n}{2}\left[a+l\right]$Sn=n2[a+l]
This formula is much simpler to work with, without the 'brackets inside the brackets' present in the first form.
For any arithmetic sequence with starting value $a$a and common difference $d$d, we can find the sum of the first $n$n terms, using:
$S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$Sn=n2(2a+(n−1)d)
OR
$S_n=\frac{n}{2}\left(a+l\right)$Sn=n2(a+l),
Where $l$l is the last term of the sequence, and:
$l=T_n=a+(n-1)d$l=Tn=a+(n−1)d
Practice questions
QUESTION 1
The first term of an arithmetic sequence is $6$6 and the $6$6th term is $26$26.
If $d$d is the difference between terms, solve for $d$d.
Hence, find the sum of the first $13$13 terms.
QUESTION 2
Consider the arithmetic sequence $4$4, $1$1, $-2$−2, …
Write a simplified expression for the sum of the first $n$n terms.
Find the sum of the sequence from the $17$17th to the $25$25th term, inclusive.
QUESTION 3
Consider the arithmetic series $-10-6-2$−10−6−2 $+$+ ... $+$+ $58$58.
If $n$n is the number of terms in the series, solve for $n$n.
Hence, find the sum of the series.
Applications of arithmetic series
Just as in the previous lesson on arithmetic sequences, questions may involve algebraic manipulation and simultaneous equations to solve for $a$a, $n$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 $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$Sn=n2(2a+(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 $T_n$Tn 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, $a$a and $d$d and use the formula $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$Sn=n2(2a+(n−1)d) to find the sum.
Do: $n=5$n=5, $a=10$a=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 $a$a, $d$d, and $S_n$Sn, substitute these values into $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$Sn=n2(2a+(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: $a=15,d=2,S_n=147$a=15,d=2,Sn=147 and hence:
| $S_n$Sn | $=$= | $\frac{n}{2}\left(2a+\left(n-1\right)d\right)$n2(2a+(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 $a=105$a=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 $a=105$a=105, $d=7$d=7 and solving $t_n=196$tn=196 for $n$n using $t_n=a+(n-1)d$tn=a+(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 $S_n=\frac{n}{2}\left(a+l\right)$Sn=n2(a+l), we have:
| $S_{14}$S14 | $=$= | $\frac{14}{2}\left(105+196\right)$142(105+196) |
| $=$= | $2100$2100 |
Practice questions
QUESTION 4
Caitlin starts training for a $4.85$4.85 km charity trail run by running every week for $28$28 weeks. She runs $1$1km of the course in the first week and each week after that she runs $350$350 metres more than the previous week, until she completes the course in one week. She then continues to run the whole course each week.
How far does she run in the $12$12th week? Give your answer correct to two decimal places, if necessary.
If she runs the full course for the first time in week $n$n, solve for $n$n.
What is the total distance that Caitlin runs in $28$28 weeks? Give your answer correct to two decimal places, if necessary.
QUESTION 5
The first term of an arithmetic sequence is $a$a and its common difference is $d$d.
Form an equation for $a$a in terms of $d$d, given that the $15$15th term is $48$48.
Form another equation for $a$a in terms of $d$d, given that the sum of the first $6$6 terms is $12$12.
Solve for the common difference, $d$d.
Hence solve for the first term, $a$a.
Hence find the sum of the first $35$35 terms of the sequence.
Simplified sums
Given an equation such as $S_n=2n^2-5n$Sn=2n2−5n, we could calculate the sum of any part of the arithmetic sequence that forms part of the series. $S_{10}$S10 would be the sum of the first ten terms. If we wanted to calculate the sum of the $11$11th to $20$20th terms of the sequence, we could do this by calculating $S_{20}-S_{10}$S20−S10. But how could we find information about the terms that form the sequence?
The best way to think about this is to consider a recursive formula for the sum of any number of terms. $S_n$Sn increases by adding in the next term: for example, $S_4$S4 is $S_3$S3 plus the fourth term $T_4$T4, and $S_{10}$S10 can be considered as $S_9$S9 plus $T_{10}$T10. We can see that the sum of any $n$n terms is equal to the sum of $n-1$n−1 terms, plus the $n$nth term. This gives the formula:
$S_n=S_{n-1}+T_n$Sn=Sn−1+Tn.
So, how can we use this form to find an expression for $T_n$Tn?
$S_n$Sn is defined as the sum of the first $n$n terms of a sequence, so, $S_1$S1 must be "the sum of the first term", that is, the first term $T_1=a$T1=a. $S_2$S2 gives us the sum of the first two terms, and so, if we calculate $S_2-S_1$S2−S1, we will find the second term $T_2$T2. Then by evaluating $T_2-T_1$T2−T1 we will find the common difference $d$d.
For our example above $S_n=2n^2-5n$Sn=2n2−5n:
$S_1=2(1)^2-5(1)=-3$S1=2(1)2−5(1)=−3 and $S_2=2(2)^2-5(2)=-2$S2=2(2)2−5(2)=−2
This means $T_2$T2 is:
$-2-(-3)=1$−2−(−3)=1
And our sequence commences with $-3,1,...$−3,1,... The common difference is $4$4. From this, we are able to use $T_n=a+(n-1)d$Tn=a+(n−1)d to find an expression for any term in the sequence:
$T_n=-3+(n-1)4=4n-7$Tn=−3+(n−1)4=4n−7
Practice question
Question 6
The sum of the first $n$n terms of an arithmetic sequence is $S_n=5n^2-10n$Sn=5n2−10n.
Find the value of $a$a, first term of the sequence.
Find the value of $T_2$T2, the second term.
Hence evaluate the common difference, $d$d.
Form a simplified expression for $T_n$Tn, the $n$nth term of the sequence.