Common Difference
Common Difference
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.
Fact:
Sequences with a common difference are called linear sequences.
Definition: 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.
\begin{align*} d& = {T}_{2}-{T}_{1} \\ & = 7-10 \\ & = -3 \end{align*}
Let us check another two terms:
\begin{align*} d& = {T}_{4}-{T}_{3} \\ & = 1-4 \\ & = -3 \end{align*}
We see that \(d\) is constant.
In general, \(d=T_{n}-T_{n-1}\)
Important:
\(d \ne T_{n-1}-T_{n}\) for example, \(d={T}_{2}-{T}_{1}\), not \({T}_{1}-{T}_{2}\).
Example
Question
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?
Figure 3.2: Two more people can be seated for each table added.
Make a table to see the pattern
|
Number of Tables, \(n\) |
Number of people seated |
Pattern |
|
\(\text{1}\) |
\(4=4\) |
\(=4+2(0)\) |
|
\(\text{2}\) |
\(4+2=6\) |
\(=4+2(1)\) |
|
\(\text{3}\) |
\(4+2+2=8\) |
\(=4+2(2)\) |
|
\(\text{4}\) |
\(4+2+2+2=10\) |
\(=4+2(3)\) |
|
\(\vdots\) |
\(\vdots\) |
\(\vdots\) |
|
n |
\(4+2+2+2+\cdots +2\) |
\(=4+2(n-1)\) |
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 \({T}_{n}=4+2(n-1)\)
Calculate the \(12^{\text{th}}\) term, in other words, find \({T}_{n}\) if \(n=12\)
\begin{align*} {T}_{12}& = 4+2(12-1) \\ & = 4+2(11) \\ & = 4 + 22 \\ & = 26 \end{align*}Therefore \(\text{26}\) people can be seated at \(\text{12}\) tables.
Calculate the number of tables needed to seat \(\text{20}\) people, in other words, find \(n\) if \({T}_{n}=20\)
\begin{align*} {T}_{n}& = 4+2(n-1) \\ 20& = 4+2(n-1) \\ 20& = 4+2n-2 \\ 20-4+2& = 2n \\ 18 & = 2n \\ \cfrac{18}{2} & = n \\ n& = 9 \end{align*}Therefore \(\text{9}\) tables are needed to seat \(\text{20}\) people.
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:
|
\(n\) |
\(\text{1}\) |
\(\text{2}\) |
\(\text{3}\) |
\(\text{4}\) |
\(\ldots\) |
|
\({T}_{n}\) |
\(\text{4}\) |
\(\text{6}\) |
\(\text{8}\) |
\(\text{10}\) |
\(\ldots\) |
Example
Question
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 \(\text{R}\,\text{520}\) 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 \({T}_{n}=120+15(n-1)\)
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:
\begin{align*} {T}_{n}& = 120+15(n-1) \\ \therefore {T}_{30}& = 120+15(30-1) \\ & = 120+15(29) \\ & = 120 + 435 \\ & = 555 \end{align*}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 \(\text{R}\,\text{520}\).
We know that:
\[{T}_{n} = 120+15(n-1)\]Therefore, if \(T_n=520= 120+15(n-1)\)
Solving for \(n\), we get:
\begin{align*} 520& = 120+15(n-1) \\ 520& = 120+15n-15 \\ 520& = 105+15n \\ 405& = 15n \\ \cfrac{405}{15} & = n \\ n& = 27 \end{align*}Therefore it is cheaper for Raymond to purchase the unlimited data plan if he uses more than \(\text{27}\) \(\text{GB}\) per month.
Learn more about number patterns:
Summary
-
The general term is expressed as the \(n^{\text{th}}\) term and is written as \({T}_{n}\).
-
We define the common difference \(d\) of a sequence as the difference between any two successive terms, where \(d=T_{n}-T(n-1)\)
-
We can work out a general formula for each number pattern and use it to determine any term in the pattern.
This lesson is part of:
Sequences and Series