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How do I test for an arithmetic sequence?
in Mathematics by Wooden (1,878 points)
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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 = \left(a+d\right)+d =a+2d \\ {T}_{4}&= {T}_{3}+d=\left(a+2d\right)+d=a+3d \\ \vdots & \quad \quad \qquad \vdots \quad \quad \qquad \vdots \quad \quad \qquad \vdots \\ {T}_{n}&= {T}_{n-1} + d =\left(a+(n-2)d\right)+d = a + \left(n-1\right)d \end{align*}

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

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

Arithmetic sequence

An arithmetic (or linear) sequence is an ordered set of numbers (called terms) in which each new term is calculated by adding a constant value to the previous term:

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

where

  • \({T}_{n}\) is the \(n\)\(^{\text{th}}\) term;

  • \(n\) is the position of the term in the sequence;

  • \(a\) is the first term;

  • \(d\) is the common difference.

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.

Worked example 1: Arithmetic sequence

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 \(\left(n \in \{1;2;3; \ldots \} \right)\) 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 &= \frac{192}{4} \\ &= 48 \\ \therefore T_{48} &= 173 \end{align*}

Write the final answer

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

by Wooden (1,878 points)

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