Part 6: Series and Sequences | Free Worksheet and Solutions

We will go through everything you need to know about series and sequences, including arithmetic progressions and geometric progressions.

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In this article, delve into series and sequences, an essential skill in high school mathematics.

Being able to successfully solve simple series and sequences skills allows students to build a solid foundation and enables them to harder problems later encountered.

This includes being able to proficiently solve questions involving exponential growth/decay, financial mathematics, etc.

 

In this article we discuss:

 

NESA Syllabus Outcomes

NESA requires students to be proficient in the following syllabus outcomes:

  • Know the difference between a sequence and a series
  • Recognise and use the recursive definition of an arithmetic sequence: \( T_{n} = T_{n – 1} + d, \ T_{1} = a \)
  • Establish and use the formula for the  \( n^{th} \) term (where \( 𝑛 \) is a positive integer) of an arithmetic sequence: \( 𝑇 = 𝑎 + (𝑛 − 1) 𝑑 \), where \( 𝑎 \) is the first term and \( 𝑑 \) is the common difference, and recognise its linear nature
  • Establish and use the formulae for the sum of the first \( 𝑛 \) terms of an arithmetic sequence:  \( S_{n} = \frac{n}{2} ( a + l )\) where \( 𝑙 \) is the last term in the sequence and \( S_{n} = \frac{n}{2} ( 2a + (n-1)d) \)
  • Identify and use arithmetic sequences and arithmetic series in contexts involving discrete linear growth or decay such as simple interest (ACMMM070)
  • Recognise and use the recursive definition of a geometric sequence:  \( T_{n} = rT_{n-1}, \ T_{1} = a \) (ACMMM072)
  • Establish and use the formula for the \( 𝑛^{th} \) term of a geometric sequence: \( T_{n} = ar^{n-1} \), where \( 𝑎 \) is the first term, \( 𝑟 \) is the common ratio and \( 𝑛 \) is a positive integer, and recognise its exponential nature (ACMMM073)
  • Establish and use the formula for the sum of the first \( 𝑛 \) terms of a geometric sequence:  \( S_{n} = \frac{a(r^{n})}{1-r} = \frac{a(r^{n} – 1)}{r-1}\)(ACMMM075)
  • Derive and use the formula for the limiting sum of a geometric series with \( |𝑟|< 1: S =\frac{a}{1-r} \)

 

Assumed Knowledge

Students should already be familiar with basic arithmetic operations and indices. This includes being able to recognise sum notation and use the basic index laws to solve for variables.

 

Introduction into Series and Sequences

Sequence

A sequence is a set of numbers that follow a certain rule. Essentially a sequence is an ordered list with the being separated by commas.

Each number in the sequence is called a term, \( T_{n} \), with \( n \) being the numerical position of the number. E.g. \( 3, 6, 9, 12, 15… \)

A sequence can be finite (i.e. the number of terms is fixed), or infinite (i.e. where the number of terms in the sequence is infinite). An infinite sequence ends with “…”.

 

Series

Conversely, a series is a sum of the terms of a sequence.

I.e. \( 1 + 2 + 3 + 4 … \) is a series, whereas \( 1, 2, 3, 4… \) is a sequence.

Each term in a series is separated by a plus symbol. The following sigma/sum notation can also be used to define a series:

\begin{align}
\sum_{k = 1}^{n} T_{k}
\end{align}

Where \( n \) refers to the number of terms and \( T_{k} \) refers to the rule/pattern followed by the series.

To find the sum of \( a \) of the first \( n \) terms of a sequence/series, the following formula can be used:

\begin{align}
S_{n} = T{1} + T_{2} + T_{3} + … + T_{n}
\end{align}

We can also rearrange this formula to make the \( n^{th} \) term of a series the subject instead:

\begin{align}
S_{n-1} &= T_{1} + T_{2} + T_{3} + … + T_{n-1} \\
S_{n} &= T_{1} + T_{2} + T_{3} + … + T_{n} \\
∴ T_{n} &= S_{n} – S_{n-1}
\end{align}

Example 1

The term of a sequence is defined by \( T_{n} = 6 + 5k – k^{2} \). Which term in the sequence has the value of 0?

Solution 1

As we are given the formula to define a term, all we need to do is equate this formula with 0:

\begin{align}
6 – 5k- k^{2} &= 0 \\
k^{2} + 5k – 6 &= 0 \\
(k-1)(k+6) &= 0 \\
k=1 \ or \ k=-6
\end{align}

However, \( k > 0 \). Therefore the 1st term \( (T_{1}) \) is 0.

 

Arithmetic Progressions (AP)

An arithmetic progression is a series or sequence where each term is determined by adding a constant to the preceding term:

Arithmetic Sequence: \( 3, 5, 7, 9… \)

Arithmetic Series: \( 3 + 5 + 7 + 9 … \)

 

Common Difference

The constant added to the preceding term is known as the common difference, \( d \)

The common difference can be positive or negative.

To find the common difference in an arithmetic progression, two consequential terms are subtracted from each other:

\begin{align}
d = T_{n} – T_{n-1}
\end{align}

 

Proving an Arithmetic Series

To prove an arithmetic series, we have to prove that the difference between two sets of consequential terms is the same. I.e.:

\begin{align}
T_{2} – T_{1} = T_{3} – T_{2}
\end{align}

 

Terms of an Arithmetic Progression

Below is the formula associated with terms in an arithmetic progression:

\begin{align}
T_{n} = a + (n-1)d
\end{align}

Where \( a \) = the first term, \( d \) = the common difference and \( n \) = the number of terms in the AP.

 

Sum of an Arithmetic Progression

There are two formulae which can be used to find the sum of an arithmetic progression:

\begin{align}
S_{n} = \frac{n}{2} \big{(} 2a + (n-1) d \big{)}
\end{align}

Where \( a \) = the first term, \( d \) = the common difference and \( n \) = the number of terms in the AP.

\begin{align}
S_{n} = \frac{n}{2} (a + l)
\end{align}

Where \( a \) = the first term, \( l \) = the last term and \( n \)= the number of terms in the AP.

 

Problems Involving Arithmetic Progressions

There are various types of problems which involve arithmetic progressions.

These include:

  • Finding the sum
  • Finding a specific term
  • Determining the common difference

Example 2

The first term of an arithmetic series is 4 and the fifth term is four times the third term. Find the common difference.

Solution 2

We know that \( a = 4 \) and \( T_{5} = 4T_{3} \). Using the formula for a term in an arithmetic progression \( \big{(} T_{n} = a + (n-1)d \big{)} \):

\begin{align}
T_{5} &= 4T_{3} \\
4 + (5-1)d &= 4 \big{(} 4 + (3-1)d \big{)} \\
4 + 4d &= 16 + 8d \\
∴ d &= 3
\end{align}

Want to make sure the next mark in your series is a Band 6?

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Geometric Progressions (GP)

A geometric progression is a series or sequence where each term is found by multiplying the previous term by a constant:

Geometric Sequence: \(  5, 25, 125, 625… \)

Geometric Series: \(  5 + 25 + 125 + 625… \)

 

Common Ratio

The constant which is multiplied to the preceding term is known as the common ratio,\( r \).

The common ratio can be positive or negative.

To find the common ratio in a geometric progression, two consequential terms are divided by each other:

\begin{align}
r = \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} = … = \frac{T_{n}}{T_{n-1}}
\end{align}

 

Terms of a Geometric Progression

Below is the formula associated with terms in a geometric progression:

\begin{align}
T_{n} =ar^{n-1}
\end{align}

Where \( a \) = the first term, \( r \) = the common ratio and \( n \) = the number of terms in the GP.

 

Sum of a Geometric Progression

Similar to Arithmetic Progressions, there is a general formula for the sum of a geometric progression. However, it varies depending on the value of \( r \):

When \( r <1\):

\begin{align}
S_{n} = \frac{a(1-r^{2})}{1-r}
\end{align}

Where \( a \) = the first term, \( r \) = the common ratio and \( n \) = the number of terms in the GP.

When \( r > 1\):

\begin{align}
S_{n} = \frac{a(r^{n} – 1)}{r-1}
\end{align}

Where \( a \) = the first term, \( r \) = the common ratio and \( n \) = the number of terms in the GP.

 

The Limiting Sum

If \( – 1 < r < 1 \) then \( S_{n} \) will approach a fixed constant value as \( n \rightarrow \infty \).

This fixed constant is also known as the limiting sum/sum to infinity \( (S_{\infty}) \) and is the total sum of that infinite geometric series.

To find it we use the following formula:

\begin{align}
S_{\infty} = \frac{a}{1-r}
\end{align}

Where \( a \) = the first term, \( r \) = the common ratio and \( -1 < r < 1 \big{(} |r| < 1 \big{)} \).

 

Problems Involving Geometric Progressions

There are various types of problems which involve geometric progressions.

These include:

  • Sum to infinity
  • Home loan calculations
  • Super calculations
  • Population growth/decline

 

Some questions can even involve a mixture of an arithmetic and geometric progressions. In these cases, we want to solve the two progressions separately.

Example 3

The cicada population in the Kakadu rainforest decreases by 15% each year. However, due to rehabilitation programs, 70 extra cicadas are introduced at the beginning of each year. There were 1800 insects at the beginning of 2010. How many insects will there be at the beginning of 2020 if this pattern continues?

Solution 3

\begin{align}
A_{2010} &= 1800 \\
A_{2011} &= 1800(0.85) + 70 \\
A_{2012} &= A_{2011} \times (0.85) + 70 \\
&= (0.85)^{2} (1800) + 70(1 + 0.85)
\end{align}

Following this pattern:

\begin{align}
A_{2020} &= (0.85)^{10}(1800) + 70 \big{(}1 + 0.85 + (0.85)^{2} + … + (0.85){9} \big{)}\\
S_{2020} &= (0.85)^{10}(1800) + 70 \frac{1 \big{(} 1-(0.85)^{10} \big{)}}{1 – 0.85} \ as \ r <1 \\
S_{2020} &= 729 \ cicadas
\end{align}

 

Concept Check Questions

1. A series is denoted by \( T_{n} = 78 – 2_{n} \). Determine \( S_{5} \).

2. The first term of an arithmetic series is \( 8 \) and the second term is half times the fifth term. Find the common difference.

3. A series is denoted by \( T_{n} = 3^{n} \). Determine \( S_{4} \).

4. Consider the series \( \frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \frac{1}{625} + … \). Find the limiting sum.

5. Express \( 0.\dot{2} \) as the simplest rational number.

 

Concept Check Solutions

1. \( 360 \)

2. \(  4 \)

3. \( 120 \)

4. \( \frac{1}{4} \)

5. \( \frac{2}{9} \)

 

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