Arithmetic Series

A series is a sequence where the goal is to add all the terms together. We will study arithmetic series and geometric series.

Recall: Notation from Sequences:

$a$ is first term

$d$ is difference, the amount we add each time

$n$ is the number of terms in the series

We will also introduce $l$ , which is the last term of the series. Since there are $n$ terms in the series and we have an $n$ th term equation for arithmetic sequences, we have a formula for $l$ :

$l=a+(n-1)d$

Make sure you are happy with the following topics before continuing.

Sequences

Adding the Terms

The sum of an arithmetic series with $n$ terms is:

$S_{n}=\dfrac{n(a+l)}{2}$

We can prove this result:

$\begin{aligned}S_{n}&=a+(a+d)+(a+2d)+...+(l-d)+l\\[1.2em]&=l+(l-d)+(l-2d)+...+(a+d)+a\\[1.2em]2S_{n}&=(a+l)+(a+l)+(a+l)+...+(a+l)+(a+l)\\[1.2em]S_{n}&=\dfrac{n(a+l)}{2}\end{aligned}$

If we substitute in our formula for $l$ , we get:

$S_{n}=\dfrac{n}{2}(2a+(n-1)d)$

Sum Notation

$\sum$ means sum, and we can use it instead of writing $S_{n}$ to represent arithmetic series.

Example: $\sum_{n=1}^{20}(3n+4)$ means the sum up to the $20$ th term of the arithmetic progression defined by $3n+4$ .

Natural Number Arithmetic Progressions

The sum of the first $n$ natural numbers (positive whole numbers) is:

$S_n = 1 + 2 + 3 + ... + (n-1) + n$

So $a=1$ , $l = n$ and $n = n$ .

If we put these values into the formula we have already seen, we would get:

$S_n = \dfrac{1}{2} n (n+1)$

Example 1: Arithmetic Series in Practice

Jon is training for a marathon. Last week his furthest run was $4$ miles. He plans to increase the length he runs by $1.1$ miles per day. What is the total distance he has run by the time he reaches his goal of $26$ miles?

[2 marks]

$a=4$

$d=1.1$

$l=26$

Find number of terms between $a$ and $l$

$n=\dfrac{26-4}{1.1}=\dfrac{22}{1.1}=20$

Need to add $1$ because both the first and last terms are included.

n=21

Substitute into formula:

$\begin{aligned}S_{n}&=\dfrac{n(a+l)}{2} \\[1.2em] S_{21}&=\dfrac{21(4+26)}{2}\\[1.2em]&= \dfrac{21\times 30}{2}\\[1.2em]&=\dfrac{630}{2}\\[1.2em]&=315 \text{ miles} \end{aligned}$

Example 2: Sum Notation

Find $\sum_{n=1}^{50}(3n+4)$

[2 marks]

This means the sum of the first $50$ terms of the sequence defined by $3n+4$ .

$a=3\times 1+4=3+4=7$

$d=3$

$n=50$

Substitute into formula:

$\begin{aligned}\sum_{n=1}^{50}(3n+4)&=\dfrac{50}{2}(2\times 7+(50-1)3)\\[1.2em]&=25(14+49\times 3)\\[1.2em]&=25(14+147)\\[1.2em]&=25\times 161\\[1.2em]&=4025\end{aligned}$

Arithmetic Series Example Questions

Question 1: Consider the sum of the first $k$ natural numbers.

i) Express this in sum notation.

ii) Find a formula for the sum in terms of $k$ .

iii) What is the value of this sum when $k=100$ ?

iv) If the sum has value $55$ , what is the value of $k$ ?

[8 marks]

i) $\sum_{n=1}^{k}n$

ii) Arithmetic progression with:

$a=1$

$d=1$

$n=k$

Sub into formula:

$\begin{aligned}\sum_{n=1}^{k}n&=\dfrac{k}{2}(2\times 1+(k-1)1)\\[1.2em]&=\dfrac{k}{2}(2+k-1)\\[1.2em]&=\dfrac{k(k+1)}{2}\end{aligned}$

iii) $k=100$

$\begin{aligned}\sum_{n=1}^{100}n&=\dfrac{100(100+1)}{2}\\[1.2em]&=\dfrac{100\times 101}{2}\\[1.2em]&=\dfrac{10100}{2}\\[1.2em]&=5050\end{aligned}$

iv) $\sum_{n=1}^{k}n=55$

$\dfrac{k(k+1)}{2}=55$

$k(k+1)=110$

$k^{2}+k=110$

$k^{2}+k-110=0$

$(k-10)(k+11)=0$

$k=10$ or $k=-11$

$-11$ not feasible

k=10

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Question 2: Find the sum of the first ten terms of the arithmetic series that begins

3,8,13,18...

[3 marks]

$a=3$

$d=5$

$n=10$

Substitute into formula:

$\begin{aligned}S_{n}&=\dfrac{10}{2}(2\times 3+(10-1)5)\\[1.2em]&=5(6+9\times 5)\\[1.2em]&=5(6+45)\\[1.2em]&=5\times 51\\[1.2em]&=255\end{aligned}$

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Question 3: At the start of every month, Jenny deposits an amount of money into her savings account. The first month she put in $£10$ , the second month $£20$ , the third month $£30$ , and so on. How much does she have saved after $2$ years?

[4 marks]

$2$ years is $24$ months.

$a=10$

$d=10$

$n=24$

Substitute into formula:

$\begin{aligned}S_{n}&=\dfrac{24}{2}(2\times 10+(24-1)10)\\[1.2em]&=12(20+23\times 10)\\[1.2em]&=12(20+230)\\[1.2em]&=12\times 250\\[1.2em]&=3000\end{aligned}$

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