This sequence has a factor of \(\frac{1}{3}\) between successive terms.

The pattern is continued by multiplying the previous term by \(\frac{1}{3}\) which is equivalent to dividing
the previous term by 3.

Some learners may see example 3 as \(2^{1}; 2^{2}; 2^{3}; \ldots\) and see a pattern with the powers. You may
choose to discuss this in class as a precursor to geometric series which will be introduced in Grade 12.

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Worked example 1: Study table

You and \(\text{3}\) friends decide to study for Maths and are sitting together at a square table. A few
minutes later, \(\text{2}\) other friends arrive and would like to sit at your table. You move another table
next to yours so that \(\text{6}\) people can sit at the table. Another \(\text{2}\) friends also want to join
your group, so you take a third table and add it to the existing tables. Now \(\text{8}\) people can sit
together.

Examine how the number of people sitting is related to the number of tables. Is there a pattern?

Make a table to see if a pattern forms

Number of tables, n

Number of people seated

\(\text{1}\)

\(4=4\)

\(\text{2}\)

\(4+2=6\)

\(\text{3}\)

\(4+2+2=8\)

\(\text{4}\)

\(4+2+2+2=10\)

\(\vdots\)

\(\vdots\)

n

\(4+2+2+2+\cdots +2\)

Describe the pattern

We can see that for \(\text{3}\) tables we can seat \(\text{8}\) people, for \(\text{4}\) tables we can seat
\(\text{10}\) people and so on. We started out with \(\text{4}\) people and added two each time. So for each
table added, the number of people increased by \(\text{2}\).

So the pattern formed is \(4; 6; 8; 10; \ldots\).

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To describe terms in a number pattern we use the following notation:

The first term of a sequence is \({T}_{1}\).

The fourth term of a sequence is \({T}_{4}\).

The tenth term of a sequence is \({T}_{10}\).

The general term is often expressed as the \(n^{\text{th}}\) term and is written as \({T}_{n}\).

A sequence does not have to follow a pattern but, when it does, we can write down the general formula to
calculate any term. For example, consider the following linear sequence:

\(1; 3; 5; 7; 9; \ldots\)

The \(n^{\text{th}}\) term is given by the general formula:

\({T}_{n}=2n-1\)

You can check this by substituting values into the formula:

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 find any term in the sequence.

Common difference (EMAZ)

Consider the following sequence:

\[6; 1; -4; -9; ...\]

We can see that each term is decreasing by 5 but how would we determine the general formula for the
\(n^{\text{th}}\) term? Let us try to do this with a table.

Term number

\(T_{1}\)

\(T_{2}\)

\(T_{3}\)

\(T_{4}\)

\({T}_{n}\)

Term

\(\text{6}\)

\(\text{1}\)

\(-\text{4}\)

\(-\text{9}\)

\({T}_{n}\)

Formula

\(6 - 0 \times 5\)

\(6 - 1 \times 5\)

\(6-2 \times 5\)

\(6-3 \times 5\)

\(6 - (n-1) \times 5\)

You can see that the difference between the successive terms is always the coefficient of \(n\) in the formula.
This is called a common difference.

Therefore, for sequences with a common difference, the general formula will always be of the form:
\(T_{n}=dn+c\) where \(d\) is the difference between each term and \(c\) is some constant.

Sequences with a common difference are called linear sequences.

Common difference

The common difference is the difference between any term and the term before it. The common difference is
denoted by \(d\).

For example, consider the sequence \(10; 7; 4; 1; \ldots\)

To calculate the common difference, we find the difference between any term and the previous term.

Let us find the common difference between the first two terms.

\(d \ne T_{n-1}-T_{n}\) for example, \(d={T}_{2}-{T}_{1}\), not \({T}_{1}-{T}_{2}\).

Worked example 2: Study table, continued

As before, you and \(\text{3}\) friends are studying for Maths and are sitting together at a square table.
A few minutes later \(\text{2}\) other friends arrive so you move another table next to yours. Now
\(\text{6}\) people can sit at the table. Another \(\text{2}\) friends also join your group, so you take a
third table and add it to the existing tables. Now \(\text{8}\) people can sit together as shown below.

Find an expression for the number of people seated at \(n\) tables.

Use the general formula to determine how many people can sit around \(\text{12}\) tables.

How many tables are needed to seat \(\text{20}\) people?

Make a table to see the pattern

Number of Tables, \(n\)

Number of people seated

Pattern

\(\text{1}\)

\(4=4\)

\(=4+2\left(0\right)\)

\(\text{2}\)

\(4+2=6\)

\(=4+2\left(1\right)\)

\(\text{3}\)

\(4+2+2=8\)

\(=4+2\left(2\right)\)

\(\text{4}\)

\(4+2+2+2=10\)

\(=4+2\left(3\right)\)

\(\vdots\)

\(\vdots\)

\(\vdots\)

n

\(4+2+2+2+\cdots +2\)

\(=4+2\left(n-1\right)\)

Note: There may be variations in how you think of the pattern in this problem. For
example, you may view this problem as the person on one end fixed, two people seated opposite each other per
table and one person at the other end fixed. This results in \(1 + 2n + 1 = 2n + 2\). Your formula for
\(T_n\) will still be correct.

Describe the pattern

The number of people seated at \(n\) tables is \({T}_{n}=4+2\left(n-1\right)\)

Calculate the \(12^{\text{th}}\) term, in other words, find \({T}_{n}\) if \(n=12\)

Therefore \(\text{9}\) tables are needed to seat \(\text{20}\) people.

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It is important to note the difference between \(n\) and \({T}_{n}\). \(n\) can be compared to a place holder
indicating the position of the term in the sequence, while \({T}_{n}\) is the value of the place held by \(n\).
From our example above, the first table holds \(\text{4}\) people. So for \(n=1\), the value of \({T}_{1}=4\)
and so on:

\(n\)

\(\text{1}\)

\(\text{2}\)

\(\text{3}\)

\(\text{4}\)

\(\ldots\)

\({T}_{n}\)

\(\text{4}\)

\(\text{6}\)

\(\text{8}\)

\(\text{10}\)

\(\ldots\)

Worked example 3: Data plans

Raymond subscribes to a limited data plan from Vodacell. The limited data plans cost
\(\text{R}\,\text{120}\) for \(\text{1}\) gigabyte (GB) per month, \(\text{R}\,\text{135}\) for \(\text{2}\)
\(\text{GB}\) per month and \(\text{R}\,\text{150}\) for \(\text{3}\) \(\text{GB}\) per month. Assume this
pattern continues indefinitely.

Use a table to set up the pattern of the cost of the data plans.

Find the general formula for the sequence.

Use the general formula to determine the cost for a \(\text{30}\) \(\text{GB}\) data plan.

The cost of an unlimited data plan is \(\text{R}\,\text{510}\) per month. Determine the amount of data
Raymond would have to use for it to be cheaper for him to sign up for the unlimited plan.

Make a table to see the pattern

Number of GB \((n)\)

\(\text{1}\)

\(\text{2}\)

\(\text{3}\)

\(\text{4}\)

Cost (in Rands)

\(\text{120}\)

\(\text{135}\)

\(\text{150}\)

\(\text{165}\)

Pattern

\(\text{120}\)

\(120+(1)(15)=135\)

\(120+(2)(15)=150\)

\(120+(3)(15)=165\)

Use the observed pattern to determine the general formula.

The price of \(n\) GB of data is \({T}_{n}=120+15\left(n-1\right)\)

Determine the cost of \(\text{30}\) \(\text{GB}\) of data.

This question requires us to determine the value of the \(30^{\text{th}}\) term, in other words, find
\({T}_{n}\) if \(n=30\). Using the general formula, we get:

Therefore the cost of a \(\text{30}\) \(\text{GB}\) data package is \(\text{R}\,\text{555}\).

Determine when it is cheaper to purchase the unlimited data plan

The final question of this worked example requires us to determine when it would be cheaper for Raymond to
purchase an unlimited data plan instead of a limited plan. In other words, we need to find \(n\) where
\(T_n\) is less than \(\text{R}\,\text{510}\).

You can see that the results are not the same - the difference is not 'common.' That means that this
sequence of numbers in not linear, and it has no common difference.

\begin{align*}
d &= T_{2}-T_{1} \text{ or } T_{3}-T_{2}\\
&= (34 b) - (30 b) \text{ or } (38 b) - (34 b) \\
&= 4 b \\
\text{Therefore } T_4 &= 42 b \\
T_5 &= 46 b \\
T_6 &= 50 b
\end{align*}

Given a pattern which starts with the numbers: \(3 \; ; \; 8 \; ; \; 13 \; ; \; 18 \; ; \; \ldots\)
determine the values of \(T_{6}\) and \(T_{9}\).

To find the two missing terms, we use the equation for the general term:
\begin{align*}
T_n & = \text{6}n\text{+3} \\
T_2 & = \text{6}(\text{2})\text{+3} \\
&= \text{15} \\
T_4 & = \text{6}(\text{4})\text{+3} \\
&= \text{27}
\end{align*}

Find the general formula for the following sequences and then find \(T_{10}\), \(T_{50}\) and
\(T_{100}\)

Consider the following list:
\[- z - 5 \; ; \; - 4 z - 5 \; ; \; - 6 z - 2 \; ; \; - 8 z - 5 \; ; \; - 10 z - 5 \; ; \; \ldots\]

Find the common difference for the terms of the list. If the sequence is not linear (if it does not
have a common difference), write “no common difference”.

\begin{align*}
d & = T_{2} - T_{1} = (- 4 z - 5) - (- z - 5) = - 3 z \\
& = T_{3} - T_{2} = (- 6 z - 2) - (- 4 z - 5) = - 2 z + 3 \\
& = T_{4} - T_{3} = (- 8 z - 5) - (- 6 z - 2) = - 2 z - 3
\end{align*}
No common difference.

If you are now told that \(z = -2\), determine the values of \(T_{1}\) and \(T_{2}\).

Find the common difference for the terms of the pattern. If the sequence is not linear (if it does
not have a common difference), write “no common difference”.

If the following terms make a linear sequence:
\[\frac{k}{3} - 1 \; ; \;- \frac{5k}{3} + 2 \; ; \;- \frac{2k}{3} + 10 \; ; \;\ldots\]
Determine the value of \(k\). If the answer is a non-integer, write the answer as a simplified
fraction.

If the following terms make a linear sequence:
\[y - \frac{3}{2} \; ; \; - y - \frac{7}{2} \; ; \; - 7 y - \frac{15}{2} \; ; \; \ldots\]
find \(y\). If the answer is a non-integer, write the answer as a simplified fraction.