# Arithmetic Series

## Arithmetic Series Revision

**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 nth 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.

**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 20th 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=21Substitute 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

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

**[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}

**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}

**Question 4: **For an arithmetic series with first term 9 and difference 4, the sum is 184. Find the number of terms.

**[3 marks]**

S_{n}=184

a=9

d=4

Substitute into formula:

\dfrac{n}{2}(2\times 9+(n-1)4)=184

n(18+4n-4)=368

n(14+4n)=368

4n^{2}+14n=368

4n^{2}+14n-368=0

2n^{2}+7n-184=0

(n-8)(2n+23)=0

n=8 or n=-\dfrac{23}{2}

n=-\dfrac{23}{2} not feasible

n=8

**Question 5:** The sum of the first k natural numbers is 325. Find the value of k.

**[2 marks]**

\dfrac{1}{2}k (k+1) = 325

k^2 + k = 650

k^2 + k - 650 = 0

(k+26)(k-25) = 0

k cannot be negative so k = 25.