Numbers: Numbers and number relationships

# Unit 6: Arithmetic sequences and series

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

• Identify and work with arithmetic progressions and sequences.
• Find the sum of an arithmetic sequence.

## What you should know

Before you start this unit, make sure you can:

## Introduction

Mathematical patterns are all around us, from the incredible beauty of sunflowers to the intricate mystery and symmetry of galaxies. A list of numbers ordered according to a rule is called a number pattern or sequence. Number patterns are all about prediction. The ability to use patterns to predict the outcomes of situations is very useful.

An important application of number sequences, which you will use, is in financial calculations of loans and investments.

Each number which makes up a sequence is called a “term”.

Look at the sequences below:

1. $\scriptsize 3;\text{ }5;\text{ }7;\text{ }9;...$
2. $\scriptsize 1;\text{ }8;\text{ }27;\text{ }64;...$
3. $\scriptsize 13;\text{ }8;\text{ }3;\text{ }-2;...$

Can you identify the pattern in each case?

Number sequences can have many different and interesting patterns. Let’s examine the patterns of the listed sequences.

Sequence A: $\scriptsize 3;\text{ }5;\text{ }7;\text{ }9;...$

The pattern is formed by adding $\scriptsize 2$ to the previous term. This type of number pattern is called a linear sequence. In a linear sequence the pattern increases (or decreases) by the same amount each time. The next five terms will be: $\scriptsize 11\text{ };\text{ }13;\text{ }15;\text{ }17;\text{ }19;...$

Sequence B: $\scriptsize 1;\text{ }8;\text{ }27;\text{ }64;...$

This sequence is formed by cubing the natural numbers. So the next five terms will be: $\scriptsize 125;\text{ 217};\text{ 343};\text{ 512};\text{ }729...$

Notice that this number pattern is not the same as the linear number pattern we saw in Sequence A, where the pattern increased by the same amount each time.

Sequence C: $\scriptsize 13;\text{ }8;\text{ }3;\text{ }-2;...$

There is a difference of $\scriptsize -5$ between consecutive terms. This is another example of a linear number pattern or sequence. The pattern is continued by adding $\scriptsize -5$ to the previous term. So, the next five terms will be: $\scriptsize -7;\text{ -12};\text{ -17};\text{ -22};\text{ }-27...$

A sequence can be infinite, as in the examples above, or it can be finite. For example, the sequence $\scriptsize \{13;\text{ }8;\text{ }3;\text{ }-2\}$ is finite and contains only four terms.

As you can see, listing all of the terms in a sequence can be cumbersome and take a long time. For example, to find the 20th term would require listing all $\scriptsize 20$ terms. A more efficient way to determine a specific term is to write a formula to define the sequence. When a sequence follows a specific pattern we can write down the general formula to calculate any term.

One type of formula is an explicit formula, which defines the terms of a sequence using their position in the sequence. Explicit formulae are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence, where n is any positive number.

In this unit we will use and discuss the arithmetic formula that gives you a quick way to find any term in a linear sequence and the position of any term in the sequence.

### Did you know?

There are special patterns all around us from the number of petals on flowers to the structure of the human body. These patterns follow a famous growth pattern known as the Fibonacci sequence. The video called “The Fibonacci Sequence: Nature’s Code” shows where the Fibonacci sequence can be seen in nature.

## Arithmetic sequences

### Activity 6.1: Explore arithmetic sequences

Time required: 20 minutes

What you need:

• A pen or pencil
• A calculator
• Blank paper or a notebook

What to do:

You and your family go to a restaurant to celebrate your birthday. The restaurant does not have a table that is long enough to seat everyone attending your birthday party. So, they have to join tables together. The restaurant only has tables that can seat four people at a time. As shown below.

1. Join two tables to the original table. How many people can be seated altogether now?
2. If the waiter joins five tables together, how many people can be seated?
3. Examine how the number of people sitting is related to the number of tables. Is there a pattern?
4. How many tables would you need to join together to seat $\scriptsize 16$ people?
5. Complete the table below to see if a pattern exists.
 Number of tables $\scriptsize (n)$ $\scriptsize 1$ $\scriptsize 2$ $\scriptsize 3$ $\scriptsize 4$ $\scriptsize 5$ Number of seated people
6. Describe the pattern in words.
7. Find a general rule or formula that relates the number of people to the number of tables.
8. Use your rule to find how many tables you need if there are $\scriptsize 30$ people coming to your birthday party.

What did you find?

1. If another table is added to the original table, six people can be seated together.

If we join two tables to the original table there will be three tables in total, which can seat eight people.
2. Five tables joined can seat $\scriptsize 12$ people.
3. Two more people can be seated for each table added.
4. Since five tables can seat $\scriptsize 12$, if we have six tables then $\scriptsize 14$ people can be seated and by joining seven tables $\scriptsize 16$ people can be seated.
5. This is the completed table. We can confirm the pattern that two more people can be seated for each table added.
 Number of tables $\scriptsize (n)$ $\scriptsize 1$ $\scriptsize 2$ $\scriptsize 3$ $\scriptsize 4$ $\scriptsize 5$ Number of seated people $\scriptsize 4$ $\scriptsize 6$ $\scriptsize 8$ $\scriptsize 10$ $\scriptsize 12$
6. For each table added, the number of people increased by two. We can see that for three tables we can seat eight people, for four tables we can seat $\scriptsize 10$ people, and so on. So, the pattern formed is $\scriptsize 4;6;8;10;12;...$ .
7. $\scriptsize 1$ table can seat $\scriptsize 4$ people.
$\scriptsize 2$ tables can seat $\scriptsize 4+1\times 2$ people.
$\scriptsize 3$ tables can seat $\scriptsize 4+2\times 2$ people.
$\scriptsize 4$ tables can seat $\scriptsize 4+3\times 2$ people.We multiply $\scriptsize 2$ by one less than the table number and add $\scriptsize 4$. Therefore, for the nth table, we will have $\scriptsize 4+(n-1)2$ people. This is called the nth term or general term. We represent the general term as $\scriptsize {{T}_{n}}$. So, we can write the general formula as $\scriptsize {{T}_{n}}=4+(n-1)\times 2$. Note: This is not the only general formula that can be formed using the information in this activity. See if you can find another rule to represent the same pattern.
8. Using the formula $\scriptsize 4+2(n-1)$ we can solve for the number of tables $\scriptsize (n)$.\scriptsize \begin{align} 4+2(n-1)& =30 \\ n-1& =26\div 2 \\ & \therefore n=13+1=14 \end{align}
You will agree that it is much quicker to use the rule to calculate the number of tables and people than to draw the tables each time.

Sequences like the one in Activity 6.1 are called arithmetic sequences or linear sequences. Arithmetic sequences are also called arithmetic progressions. We add a specific, constant number to each term in the sequence to find the next term in the sequence. The number we need to add to each term to create the next term is called the common difference and is represented by the letter $\scriptsize d$. The common difference is constant throughout an arithmetic sequence because the difference between any two successive terms is always the same.

### Describing an arithmetic sequence

To describe terms in a number pattern we use the following notation: $\scriptsize {{T}_{1}};\text{ }{{T}_{2}};\text{ }{{T}_{3}};...{{T}_{n}}$

The subscript tells us the position of the term in the sequence.

$\scriptsize {{T}_{1}}$ is the first term of the sequence and is often represented using the letter $\scriptsize a$.

$\scriptsize {{T}_{2}}$ is the second term of the sequence.

$\scriptsize {{T}_{3}}$ is the third term of the sequence.

$\scriptsize {{T}_{n}}$ is the nth term or general term of the sequence.

We know that if we find the relationship between the position of a term and its value, we can find a general formula which matches the pattern and any term in the sequence.

In Activity 6.1 we discovered that the general term of $\scriptsize 4;6;8;10;12;...$ is $\scriptsize {{T}_{n}}=4+(n-1)\times 2$.

$\scriptsize {{T}_{1}}$ is added to $\scriptsize (n-1)$ and multiplied by the common difference $\scriptsize d$ to find any term in the sequence.

For sequences with a common difference, the general formula will always be: $\scriptsize {{T}_{n}}=a+(n-1)d$, where $\scriptsize a$ is the first term and $\scriptsize ~d$ is the common difference.

The common difference is the difference between any two consecutive terms in an arithmetic sequence. The common difference is denoted by $\scriptsize ~d$.

A sequence is arithmetic if there is a constant difference between the terms in the sequence. To check if a sequence is arithmetic find the difference between consecutive terms. For a sequence to be arithmetic then $\scriptsize {{T}_{2}}-{{T}_{1}}={{T}_{3}}-{{T}_{2}}$.

### Example 1

1. Write down the next three terms in the following sequence. Then decide if the sequence is arithmetic or not. If the sequence is arithmetic, write down the general term.
$\scriptsize -1;~\text{ }-4;~\text{ }-7,~\ldots$
2. The general term of an arithmetic sequence is $\scriptsize {{T}_{n}}=5+\left( n-1 \right)4$.
1. Write down $\scriptsize {{T}_{1}}$.
2. Find $\scriptsize {{T}_{3}}$ and $\scriptsize {{T}_{15}}$.
3. Find the common difference.
4. Which term has a value of $\scriptsize 85$?

Solution

1. We need to add $\scriptsize -3$ to each term to find the next term in the sequence. The next three terms are $\scriptsize -10;~\text{ }-13;~\text{ }-16,~\ldots$
\scriptsize \begin{align} & {{T}_{2}}-{{T}_{1}}=-4-(-1)=-3 \\ & {{T}_{3}}-{{T}_{2}}=-7-(-4)=-3 \\ \end{align}
Therefore, this is an arithmetic sequence with a common difference of $\scriptsize -3$.
The first term $\scriptsize a$ has a value of $\scriptsize -1$.
\scriptsize \begin{align} {{T}_{n}}& =a+(n-1)d \\ & =-1+(n-1)(-3) \\ & =-1-3n+3 \\ & =2-3n \end{align}
2. .
1. To find any term in the sequence, replace $\scriptsize n$ with the term’s position.
Replace $\scriptsize n$ with $\scriptsize 1$
$\scriptsize {{T}_{n}}=5+\left( n-1 \right)4$
\scriptsize \begin{align} {{T}_{1}}& =5+\left( 1-1 \right)4 \\ & =5\\ \end{align}
2. Replace $\scriptsize n$ with $\scriptsize 3$
\scriptsize \begin{align} {{T}_{3}}& =5+\left( 3-1 \right)4 \\ & =13\\ \end{align}
AND Replace $\scriptsize n$ with $\scriptsize 15$
\scriptsize \begin{align} {{T}_{15}}& =5+\left( 15-1 \right)4 \\ & =61\\ \end{align}
3. You need the value of two consecutive terms, so you need to find $\scriptsize {{T}_{2}}$ to find the common difference. You are told the sequence is arithmetic so you can choose any two terms to work with. We already know that $\scriptsize {{T}_{1}}=5$, so calculate $\scriptsize {{T}_{2}}$.
\scriptsize \begin{align} {{T}_{2}}& =5+\left( 2-1 \right)4 \\ & =9\\ \end{align}
So to calculate the common difference:
\scriptsize \begin{align} {{T}_{2}}-{{T}_{1}}& =9-5 \\ & =4\\ \end{align}
The common difference is $\scriptsize 4$
4. We are given the value of the term and need to calculate the term’s position. In other words, we need to find $\scriptsize n$.
\scriptsize \begin{align} 5+\left( n-1 \right)4& =85 \\ (n-1)4& =80 \\ n-1& =20 \\ n& =21\\ \end{align}
Note: The position of a term in a sequence will always be a natural number. Therefore, $\scriptsize n$ will never be a fraction or negative.

### Take note!

It is important to note the difference between $\scriptsize n$ and $\scriptsize {{T}_{n}}$. Think of $\scriptsize n$ as a place holder indicating the position of the term in the sequence, while $\scriptsize {{T}_{n}}$ is the value of the place held by $\scriptsize n$.

Work through this exercise to assess your understanding.

### Exercise 6.1

1. List the first four terms of an arithmetic sequence with the following properties:
1. $\scriptsize a=16$ and $\scriptsize d=-2$
2. $\scriptsize a=-5$ and $\scriptsize d=3$
2. Find the general term for the sequence: $\scriptsize 120;\text{ }135;\text{ 150;}...$ .
3. Find the number of terms in the arithmetic sequence: $\scriptsize \{8;\text{ }1;\text{ }-6;...;-41\}$.
4. Find the general term of the arithmetic sequence with $\scriptsize {{T}_{2}}=11$ and $\scriptsize {{T}_{5}}=32$.

The full solutions are at the end of the unit.

## Arithmetic series

Let’s say you decide to save money and start immediately by saving $\scriptsize \text{R50}$. If you increase your savings by $\scriptsize \text{R20}$ every week, how much would you have at the end of three weeks if you do not put the money in a bank account?

It is a simple process of adding your savings amounts over the period to see how much you have altogether. Since the money is not invested in a bank account it will gain no interest and only grow by the amount you add each week. So, at the end of three weeks you will have $\scriptsize \text{R}50+\text{R7}0+\text{R9}0+\text{R11}0=\text{R320}$.

If you need to calculate the total amount saved after a few weeks, for example $\scriptsize 33$ weeks, it will be quite tedious to write out and add all of those amounts manually. Can you find a quicker way to calculate how much is saved in total over a period of time?

We could write the amount saved as a sequence of terms.

$\scriptsize 50;\text{ }70\text{; }90;\text{ }110;...$

We see that this is an arithmetic sequence with a common difference of $\scriptsize 20$. But, now that we know this is an arithmetic sequence, is there a quicker way to find the sum of the terms? The sum of the terms of any sequence is called a series. There are two types of series; finite series and infinite series. In this unit, and in Maths Level 2, you will only deal with finite series.

The symbol $\scriptsize {{S}_{n}}$ is used to show the sum of the first $\scriptsize n$ terms of a finite sequence $\scriptsize {{S}_{n}}={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{n}}$.

For the sequence of savings $\scriptsize 50;\text{ }70\text{; }90;\text{ }110$ the sum of the first four terms is $\scriptsize {{S}_{4}}=50+70+90+110=320$. However, as we noted before, it will be very time-consuming to keep adding these terms together for $\scriptsize n$ terms when $\scriptsize n$ becomes large.

Since $\scriptsize {{S}_{n}}={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{n}}$ and we are dealing with an arithmetic sequence, it is possible to derive a general formula for the sum of $\scriptsize n$ terms.

The general formula to find the sum of $\scriptsize n$ terms in an arithmetic series is:
$\scriptsize {{S}_{n}}=\displaystyle \frac{n}{2}\left[ 2a+(n-1)d \right]$

You do not need to know how this formula is derived but you do need to know how to apply it.

To find the sum of your savings $\scriptsize 50;\text{ }70\text{; }90;\text{ }110;...$ after $\scriptsize 33$ weeks, substitute the values $\scriptsize a=50;\text{ }n=33;\text{ }d=20$ into the formula.

\scriptsize \begin{align} {{S}_{33}}& =\displaystyle \frac{33}{2}\left[ 2(33)+(33-1)\times 20 \right] \\ & =\displaystyle \frac{33}{2}[706] \\ & =\text{R}11\text{ 649} \end{align}

This demonstrates how useful the general formula for determining an arithmetic series is, especially when the series has many terms.

### Example 2

Find the sum of the series $\scriptsize 20+15+10+...+(-50)$.

Solution

Step $\scriptsize 1$: Determine if the series is arithmetic.

\scriptsize \begin{align} & {{T}_{3}}-{{T}_{2}}=10-15=-5 \\ & {{T}_{2}}-{{T}_{1}}=15-20=-5 \\ \end{align}

Therefore, the series is arithmetic with $\scriptsize d=-5$ and $\scriptsize a=20$.

Step $\scriptsize 2$: Calculate the value of $\scriptsize n$.

We also need to know the value of $\scriptsize n$ in $\scriptsize {{S}_{n}}=\displaystyle \frac{n}{2}\left[ 2a+(n-1)d \right]$.

We can use $\scriptsize {{T}_{n}}=a+(n-1)d$ to find $\scriptsize n$.

The last term $\scriptsize {{T}_{n}}=-50$, $\scriptsize d=-5$ and $\scriptsize a=20$.

\scriptsize \begin{align} -50& =20+(n-1)(-5) \\ -50-20& =-5(n-1) \\ \displaystyle \frac{-70}{-5}& =n-1 \\ 14+1& =n \\ 15& =n \end{align}

Step $\scriptsize 3$: Use $\scriptsize {{S}_{n}}=\displaystyle \frac{n}{2}\left[ 2a+(n-1)d \right]$ to find the sum.

\scriptsize \begin{align} {{S}_{15}}&=\displaystyle \frac{15}{2}\left[ 2(20)+(15-1)(-5) \right] \\ & =\displaystyle \frac{15}{2}\left[ -30 \right] \\ & =-225 \end{align}

## Sigma notation

Sigma notation is a compact way to write the sum of a given number of terms in a series. The symbol $\scriptsize \Sigma$ (sigma), which is the Greek capital letter S, is used to represent the sum.

The number of terms in the series = upper bound – lower bound + $\scriptsize 1$. So, the number of terms in this series is $\scriptsize 6-2+1=5$.

If you were to list the terms of the series from $\scriptsize n=2$ up to and including $\scriptsize n=6$ you would get $\scriptsize 4,6,8,10,12$. And if you count the number of terms you will have $\scriptsize 5$ terms in the series.

The given notation asks us to find the sum of the terms in the series $\scriptsize {{T}_{n}}=2n$ for $\scriptsize n=2$ through to $\scriptsize n=6$.

### Example 3

Find the value of the series $\scriptsize \sum\limits_{k=2}^{15}{2k}$.

Solution

Step $\scriptsize 1$: Find the first three terms of the series and determine if the series is arithmetic.

\scriptsize \begin{align} & {{T}_{1}}=2(1)=2 \\ & {{T}_{2}}=2(2)=4 \\ & {{T}_{3}}=2(3)=6 \\ \end{align}

We see that this is an arithmetic sequence.

\scriptsize \begin{align} & {{T}_{3}}-{{T}_{2}}=6-4=2 \\ & {{T}_{2}}-{{T}_{1}}=4-2=2 \\ & d=2 \\ \end{align}

Step $\scriptsize 2$: Work out the number of terms in the series.

Number of terms: $\scriptsize 15-2+1=14$

Step $\scriptsize 3$: Work out the sum of the series using $\scriptsize {{S}_{n}}=\displaystyle \frac{n}{2}\left[ 2a+(n-1)d \right]$.

$\scriptsize a=2;\text{ }d=2;\text{ }n=14$

\scriptsize \begin{align} {{S}_{14}}& =\displaystyle \frac{14}{2}\left[ 2(2)+(14-1)(2) \right] \\ & =7[30] \\ & =210 \end{align}

$\scriptsize \therefore \sum\limits_{k=2}^{15}{2k}=210$

Now, do this exercise to check your understanding of arithmetic series and sigma notation.

### Exercise 6.2

1. Find the sum of $\scriptsize \text{13 + 21 + 29 + }...+ 69$.
2. Jackson has minor surgery on his leg. A month after the surgery he can walk $\scriptsize 1.5\text{ km}$. Every month thereafter he walks an additional $\scriptsize 0.5\text{ km}$. After $\scriptsize 5$ weeks, what is the total number of kilometres he has walked?
3. Write the following series in sigma notation $\scriptsize \text{31 + 24 + 17 + 10 + 3}$.
4. Determine the value of $\scriptsize k$ in the series $\scriptsize \sum\limits_{n=1}^{k}{(-2n)=-20}$.
5. The sum to n terms of an arithmetic series is $\scriptsize {{S}_{n}}=\displaystyle \frac{n}{2}(7n+15)$. How many terms of the series must be added to give a sum of $\scriptsize 425$?

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to determine if a sequence is arithmetic.
• How to find the common difference.
• The formula for n terms of an arithmetic sequence.
• How to find the sum of an arithmetic series by using the summing formula.
• The definition of sigma notation and how to interpret its meaning.

# Unit 6: Assessment

#### Suggested time to complete: 40 minutes

1. Take a look at the sequence $\scriptsize 2;\text{ }6\text{; }10;\text{ }14;...$.
2. Calculate $\scriptsize {{T}_{22}}$.
3. Which term has a value of $\scriptsize 322$?
2. The $\scriptsize {{10}^{\text{th}}}$ term of an arithmetic sequence is $\scriptsize 34$ and the $\scriptsize {{16}^{\text{th}}}$ term is $\scriptsize 52$. Calculate the sum of the first $\scriptsize 16$ terms.
3. During her June holidays Siya plays a PlayStation game. She has no memory card and has to start the game from the beginning each time she plays. At the end of the first day she reaches stage $\scriptsize 5$ of the game. After restarting the game on the second day she reaches stage $\scriptsize 9$. At the end of the third day she reaches stage $\scriptsize 13$ and at the end of the fourth day she reaches stage $\scriptsize 17$.
1. If she continues with this sequence, what stage will she reach at the end of the $\scriptsize {{22}^{\text{nd}}}$ day?
2. If the game has $\scriptsize 102$ stages, on which day will she complete the game?
3. If Siya spends $\scriptsize ~5$ minutes of the first day reading and increases her reading time by $\scriptsize 3$ minutes each day, how many minutes would she spend on reading if her holiday is $\scriptsize 36$ days long.
4. To build a wall a builder finds that he will need $\scriptsize 25$ bricks in the bottom row, $\scriptsize 30$ bricks in the second row, $\scriptsize 35$ bricks in the third row, and so the pattern continues up to the $\scriptsize {{20}^{\text{th}}}$ row.
1. How many bricks are there in the last row?
2. How many bricks are there in the wall altogether?

The full solutions are at the end of the unit.

# Unit 6: Solutions

### Exercise 6.1

1. .
1. $\scriptsize a=16$ and $\scriptsize d=-2$
$\scriptsize {{T}_{1}}=a=16$
$\scriptsize {{T}_{2}}=16+1\times (-2)=14$
$\scriptsize {{T}_{3}}=16+2\times (-2)=16-4=12$
$\scriptsize {{T}_{4}}=16+3\times (-2)=16-6=10$
The first four terms are: $\scriptsize 16;\text{ }14;\text{ }12;\text{ }10$
2. $\scriptsize a=-5$ and $\scriptsize d=3$
The first four terms are: $\scriptsize -5;\text{ -2};\text{ 1};\text{ 4}$.
2. For $\scriptsize 120;\text{ }135;\text{ 150;}...$ $\scriptsize a=120;\text{ }d=135-120=150-135=15$
\scriptsize \begin{align} {{T}_{n}}& =a+(n-1)d \\ & =120+(n-1)(15) \\ & =120+15n-15 \\ & =105+15n \end{align}
3. In the sequence $\scriptsize \{8;\text{ }1;\text{ }-6;...;-41\}$, $\scriptsize a=8;\text{ }d=-7;\text{ }{{\text{T}}_{n}}=-41$
We need to solve for $\scriptsize n$.
\scriptsize \begin{align} {{\text{T}}_{n}}& =-41 \\ a+(n-1)d& =-41 \\ 8+(n-1)(-7)& =-41 \\ (n-1)(-7)& =-49 \\ n& =\displaystyle \frac{-49}{-7}+1 \\ & \therefore n=8 \end{align}
4. To find the general term of the arithmetic sequence with $\scriptsize {{T}_{2}}=11$ and $\scriptsize {{T}_{5}}=32$.
\scriptsize \begin{align} & {{T}_{2}}=11 \\ & \therefore a+d=11\text{ Eqn }1 \\ & {{T}_{5}}=32 \\ & \therefore a+4d=32\text{ Eqn }2 \\ \end{align}
Subtract Equation $\scriptsize 1$ from Equation $\scriptsize 2$ and solve simultaneously.
\scriptsize \begin{align} & a+4d=32\text{ } \\ & \underline{-a-d=-11\text{ }} \\ & \text{ 3d=21} \\ & \text{ }\therefore \text{d=7} \\ \end{align}
Substitute back into Equation $\scriptsize 1$ (or $\scriptsize 2$).
\scriptsize \begin{align} & a+7=11 \\ & \therefore a=4 \\ \end{align}
\scriptsize \begin{align} & \therefore {{T}_{n}}=4+7(n-1) \\ & =7n-3 \end{align}

Back to Exercise 6.1

### Exercise 6.2

1. Check for arithmetic progression.
$\scriptsize \text{13 + 21 + 29 + }...+69$
\scriptsize \begin{align} & {{T}_{3}}-{{T}_{2}}=29-21=8 \\ & {{T}_{2}}-{{T}_{1}}=21-13=8 \\ & \therefore d=8 \\ \end{align}
\scriptsize \begin{align} & a=13 \\ & {{T}_{n}}=69 \\ & 13+(n-1)8=69 \\ & n=\left( \displaystyle \frac{69-13}{8} \right)+1 \\ & \therefore n=8 \\ \end{align}
\scriptsize \begin{align} {{S}_{n}}& =\displaystyle \frac{8}{2}\left[ 2(13)+(8-1)9 \right] \\ & =4(89) \\ & =356 \end{align}
2. .Set up the sequence.
$\scriptsize 1.5;\text{ }2;\text{ }2.5;...\text{ }$
Sequence is arithmetic with $\scriptsize d=0.5$
\scriptsize \begin{align} {{S}_{5}}& =\displaystyle \frac{5}{2}\left[ 2(1.5)+(5-1)0.5 \right] \\ & =\displaystyle \frac{5}{2}(5) \\ & =12.5\text{ km} \end{align}
After $\scriptsize 5$ weeks, he has walked $\scriptsize 12.5\text{ km}$ in total.
3. .First determine if the series is arithmetic: $\scriptsize \text{31 + 24 + 17 + 10 + 3}$
\scriptsize \begin{align} & {{T}_{3}}-{{T}_{2}}=17-24=-7 \\ & {{T}_{2}}-{{T}_{1}}=24-31=-7 \\ & \therefore d=-7 \\ \end{align}
Then determine the general formula of the series.
\scriptsize \begin{align} {{T}_{n}}& =a+(n-1)d \\ & =31+(n-1)(-7) \\ & =38-7n \end{align}
Lastly, determine the sum of the series and write it in sigma notation.
\scriptsize \begin{align} & {{S}_{n}}=31+24+17+10+3=85 \\ & \therefore \sum\limits_{n=1}^{5}{(-7n+38)=85} \\ \end{align}
4. .$\scriptsize \sum\limits_{n=1}^{k}{(-2n)=-20}$
\scriptsize \begin{align} {{T}_{1}}& =-2(1)=-2 \\ {{T}_{2}}& =-2(2)=-4 \\ {{T}_{3}}& =-2(3)=-6 \\ \end{align}
The series is arithmetic with a common difference of $\scriptsize -2$.
\scriptsize \begin{align} {{S}_{k}}& =-20 \\ \displaystyle \frac{k}{2}\left[ 2(-2)+(k-1)(-2) \right]& =-20 \\ k\left[ -2-2k \right]& =-40 & & \text{ multiply both sides by }2 \\ -2k-2{{k}^{2}}+40& =0 & &\text{ Divide by}-2 \text{ and rewrite in standard form} \\ {{k}^{2}}+k-20& =0 \\ (k-4)(k+5)& =0 \\ k=4\text{ or }k& =-5\text{ n/a} \\ \therefore k& =4 \end{align}
5. .
\scriptsize \begin{align} & {{S}_{n}}=\displaystyle \frac{n}{2}(7n+15) \\ & {{S}_{n}}=425 \\ \end{align}
Solve for n.
\scriptsize \begin{align} \displaystyle \frac{n}{2}(7n+15) & =425 \\ n(7n+15) & =850 \\ 7{{n}^{2}}+15n-850 & =0 \\ (7n+85)(n-10) & =0 \\ \therefore \ n & =10 \end{align}
Therefore, ten terms must be added to give a sum of $\scriptsize 425$.

Back to Exercise 6.2

### Unit 6: Assessment

1. .
1. .
\scriptsize \begin{align} & 2;\text{ }6\text{; }10;\text{ }14;... \\ & {{T}_{3}}-{{T}_{2}}=10-6=4 \\ & {{T}_{2}}-{{T}_{1}}=6-2=4 \\ \end{align}
Sequence is arithmetic with $\scriptsize \text{ }d=4$
2. .
\scriptsize \begin{align} {{T}_{22}}& =2+(22-1)4 \\ & =86 \end{align}
3. .
\scriptsize \begin{align} {{T}_{n}}& =2+(n-1)4 \\ & =4n-2 \end{align}
\scriptsize \begin{align} \therefore 4n-2 & =322 \\ 4n & =324 \\ n & =81 \end{align}
$\scriptsize \therefore {{T}_{81}}=322$
2. .
$\scriptsize {{T}_{10}}=34;\text{ }{{T}_{16}}=52\text{ }$
\scriptsize \begin{align} & a+9d=34\text{ } \\ & \therefore a=34-9d\text{ Eqn }1 \\ & a+15d=52\text{ Eqn }2 \\ \end{align}
Substitute Equation 1 into Equation 2.
\scriptsize \begin{align} (34-9d)+15d & =52\text{ } \\ 34+6d & =52 \\ 6d & =18 \\ \therefore d & =3 \end{align}
Substitute back into Equation 1
$\scriptsize \therefore a=34-9(3)=7$
\scriptsize \begin{align} {{S}_{16}}& =\displaystyle \frac{16}{2}\left[ 2(7)+(15)(3) \right] \\ & =8(59) \\ & =472 \end{align}
3. .
1. .
\scriptsize \begin{align} & 5;\text{ }9;\text{ }13;\text{ }17;... \\ & 13-9=4 \\ & 9-5=4 \\ & d=4 \\ \end{align}
\scriptsize \begin{align} {{T}_{22}}& =5+(22-1)(4) \\ & =89 \end{align}
She will reach stage $\scriptsize 89$ by end of the $\scriptsize {{22}^{\text{nd}}}$ day.
2. .
\scriptsize \begin{align} {{T}_{n}}& =102 \\ 5+(n-1)(4)& =102 \\ n& =\displaystyle \frac{\left( 102-5 \right)}{4}+1 \\ & =98 \end{align}
She will complete the game on day $\scriptsize 98$.
3. .
\scriptsize \begin{align} & 5;\text{ 8};\text{ }11;... \\ & d=3 \\ \end{align}
\scriptsize \begin{align} {{S}_{36}}& =\displaystyle \frac{36}{2}\left[ 2(5)+(36-1)(3) \right] \\ & =18(115) \\ & =2070 \end{align}
She spends $\scriptsize 2070$ minutes reading.
4. .
1. How many bricks are there in the last row?
\scriptsize \begin{align} & 25;\text{ }30;\text{ }35;... \\ & a=25;\text{ }d=5 \\ \end{align}
\scriptsize \begin{align} {{T}_{20}}& =25+(20-1)(5) \\ & =120 \end{align}
There are $\scriptsize 120$ bricks in the last row.
2. .
\scriptsize \begin{align} {{S}_{20}}& =\displaystyle \frac{20}{2}\left[ 2(25)+(19)(5) \right] \\ & =10[145] \\ & =1450 \end{align}
There are $\scriptsize 1450$ bricks in the wall.

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