The General Term For An Arithmetic Sequence

The General Term For An Arithmetic Sequence

For a general arithmetic sequence with first term \(a\) and a common difference \(d\), we can generate the following terms:

\begin{align*} {T}_{1}&= a \\ {T}_{2}&= {T}_{1}+d = a + d \\ {T}_{3}&= {T}_{2}+d = (a+d)+d =a+2d \\ {T}_{4}&= {T}_{3}+d=(a+2d)+d=a+3d \\ \vdots & \quad \quad \qquad \vdots \quad \quad \qquad \vdots \quad \quad \qquad \vdots \\ {T}_{n}&= {T}_{n-1} + d =(a+(n-2)d)+d = a + (n-1)d \end{align*}

Therefore, the general formula for the \(n\)\(^{\text{th}}\) term of an arithmetic sequence is:

\[{T}_{n}= a + (n-1)d\]

Test for an arithmetic sequence

To test whether a sequence is an arithmetic sequence or not, check if the difference between any two consecutive terms is constant:

\[d = {T}_{2}-{T}_{1}={T}_{3}-{T}_{2}= \ldots = {T}_{n}-{T}_{n-1}\]

If this is not true, then the sequence is not an arithmetic sequence.

Example

Question

Given the sequence \(-15; -11; -7; \ldots 173\).

1. Is this an arithmetic sequence?
2. Find the formula of the general term.
3. Determine the number of terms in the sequence.

Check if there is a common difference between successive terms

\begin{align*} T_{2} - T_{1} &= -11 - (-15) = 4 \\ T_{3} - T_{2} &= -7 - (-11) = 4 \\ \therefore \text{This is an } & \text{arithmetic sequence with } d = 4 \end{align*}

Determine the formula for the general term

Write down the formula and the known values:

\[T_{n} = a + (n-1)d\]\[a = -15; \qquad d = 4\]\begin{align*} T_{n} &= a + (n-1)d \\ &= -15 + (n-1)(4) \\ &= -15 + 4n - 4 \\ &= 4n - 19 \end{align*}

A graph was not required for this question but it has been included to show that the points of the arithmetic sequence lie in a straight line.

Note: The numbers of the sequence are natural numbers \((n \in \{1;2;3; \ldots \} )\) and therefore we should not connect the plotted points. In the diagram above, a dotted line has been used to show that the graph of the sequence lies on a straight line.

Determine the number of terms in the sequence

\begin{align*} T_{n} &= a + (n-1)d \\ 173 &= 4n - 19 \\ 192 &= 4n \\ \therefore n &= \cfrac{192}{4} \\ &= 48 \\ \therefore T_{48} &= 173 \end{align*}

Write the final answer

Therefore, there are \(\text{48}\) terms in the sequence.

Arithmetic mean

The arithmetic mean between two numbers is the number half-way between the two numbers. In other words, it is the average of the two numbers. The arithmetic mean and the two terms form an arithmetic sequence.

For example, the arithmetic mean between \(\text{7}\) and \(\text{17}\) is calculated:

\begin{align*} \text{Arithmetic mean } &= \cfrac{7 + 17}{2} \\ &= 12 \\ \therefore 7; 12; 17 & \text{ is an arithmetic sequence} \\ T_{2} - T_{1} &= 12 - 7 = 5 \\ T_{3} - T_{2} &= 17-12 = 5 \end{align*}

Plotting a graph of the terms of a sequence sometimes helps in determining the type of sequence involved.For an arithmetic sequence, plotting \({T}_{n}\) vs. \(n\) results in the following graph:

• If the sequence is arithmetic, the plotted points will lie in a straight line.
• Arithmetic sequences are also called linear sequences, where the common difference (\(d\)) is the gradient of the straight line.

\begin{align*} T_{n} &= a + (n-1)d \\ \text{can be written as } T_{n}&= d(n - 1) + a \\ \text{which is of the } & \text{same form as } y = mx + c \end{align*}

The general formula for the \(n^{\text{th}}\) term of a quadratic sequence is: \[{T}_{n}=a{n}^{2}+bn+c\]

It is important to note that the first differences of a quadratic sequence form an arithmetic sequence. This sequence has a common difference of \(2a\) between consecutive terms. In other words, a linear sequence results from taking the first differences of a quadratic sequence.

Example

Question

Consider the pattern of white and blue blocks in the diagram below.

1. Determine the sequence formed by the white blocks \((w)\).
2. Find the sequence formed by the blue blocks \((b)\).

Pattern number \((n)\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)	\(\text{6}\)	\(n\)
No. of white blocks \((w)\)
Common difference \((d)\)

Pattern number \((n)\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)	\(\text{6}\)	\(n\)
No. of blue blocks \((b)\)
Common difference \((d)\)

Use the diagram to complete the table for the white blocks

Pattern number \((n)\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)	\(\text{6}\)	\(n\)
No. of white blocks \((w)\)	\(\text{4}\)	\(\text{8}\)	\(\text{12}\)	\(\text{16}\)	\(\text{20}\)	\(\text{24}\)	\(4n\)
Common difference \((d)\)		\(\text{4}\)	\(\text{4}\)	\(\text{4}\)	\(\text{4}\)	\(\text{4}\)

We see that the next term in the sequence is obtained by adding \(\text{4}\) to the previous term, therefore the sequence is linear and the common difference \((d)\) is \(\text{4}\).

The general term is:

\begin{align*} T_{n} &= a + (n-1)d \\ &= 4 + (n-1)(4) \\ &= 4 + 4n - 4 \\ &= 4n \end{align*}

Use the diagram to complete the table for the blue blocks

Pattern number \((n)\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)	\(\text{6}\)
No. of blue blocks \((b)\)	\(\text{0}\)	\(\text{1}\)	\(\text{4}\)	\(\text{9}\)	\(\text{16}\)	\(\text{25}\)
Difference		\(\text{1}\)	\(\text{3}\)	\(\text{5}\)	\(\text{7}\)	\(\text{9}\)

We notice that there is no common difference between successive terms. However, there is a pattern and on further investigation we see that this is in fact a quadratic sequence:

Pattern number \((n)\)	\(\text{1}\)	\(\text{2}\)	\(\text{3}\)	\(\text{4}\)	\(\text{5}\)	\(\text{6}\)	\(n\)
No. of blue blocks \((b)\)	\(\text{0}\)	\(\text{1}\)	\(\text{4}\)	\(\text{9}\)	\(\text{16}\)	\(\text{25}\)	\((n-1)^{2}\)
First difference	\(-\)	\(\text{1}\)	\(\text{3}\)	\(\text{5}\)	\(\text{7}\)	\(\text{9}\)	\(-\)
Second difference	\(-\)	\(-\)	\(\text{2}\)	\(\text{2}\)	\(\text{2}\)	\(\text{2}\)	\(-\)
Pattern	\((1-1)^{2}\)	\((2-1)^{2}\)	\((3-1)^{2}\)	\((4-1)^{2}\)	\((5-1)^{2}\)	\((6-1)^{2}\)	\((n-1)^{2}\)

\[T_{n} = (n - 1)^{2}\]

Draw a graph of \(T_{n}\) vs. \(n\) for each sequence

\begin{align*} \text{White blocks: } T_{n} &= 4n \\ \text{Blue blocks: } T_{n} &= (n-1)^{2} \\ &= n^{2} - 2n + 1 \end{align*}

Since the numbers of the sequences are natural numbers \((n \in \{1;2;3; \ldots \} )\), we should not connect the plotted points. In the diagram above, a dotted line has been used to show that the graph of the sequence formed by the white blocks \((w)\) is a straight line and the graph of the sequence formed by the blue blocks \((b)\) is a parabola.