The General Term of an Arithmetic Sequence


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The first term in a sequence of real numbers is \(a\).

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Try This Revisited

The first term in a sequence of real numbers is \(a\). Each subsequent term is calculated by adding \(d\) to the previous term. Write the first \(4\) terms of this sequence and determine the general term.

First four terms:

\(a,\)\(~a+d,\)\(~a+d+d,\)\(~a+d+d+d\)

 

The General Term of an Arithmetic Sequence

We will use this formula frequently throughout this lesson.

 

Intuitive Understanding of ​​​​\(t_n=a+(n-1)d\) ​​​​

Formulas are best remembered when they make sense.

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Example 4

Determine and graph the first five terms of \(t_n=-8+4(n-1)\).

 

 

Example 4 Continued

Determine and graph the first five terms of \(t_n=-8+4(n-1)\).

 

 

 

Example 4 Continued

Determine and graph the first five terms of \(t_n=-8+4(n-1)\).

Solution

Graph the sequence \(-8,-4,~0,~4,~8.\)

 

Arithmetic Sequences and Linear Functions

When we previously graphed the terms of an arithmetic sequence, the points were on a straight line.

 

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Example 5

Determine if the given equation represents the general term of an arithmetic sequence. For those that are arithmetic sequences, determine the first term, \(a\), and the common difference, \(d\).

  1. \(t_n=n^2\)
  2. \(t_n=7+3(n-1)\)
  3. \(t_n=6n-2\)
  4. \(t_n=\dfrac3n\)
  5. \(t_n=-4\)

Solution

Recall that an arithmetic sequence is equivalent to a linear function with its domain restricted to the natural numbers. Linear functions are given by equations that can be written in the form \(y=mx+b\).

  1. \(t_n=n^2\) is quadratic, not linear, so it is not the general term of an arithmetic sequence.
  2. \(t_n=7+3(n-1)\) is linear so it is the general term of an arithmetic sequence. The general term of an arithmetic sequence can be written as \(t_n=a+(n-1)d\); the general term we are given can be written as \(t_n=7+(n-1)(3)\).
    Therefore, the sequence has \(a=7\) and \(d=3\).
  3.  \(t_n=6n-2\) is linear so it is the general term of an arithmetic sequence.
    The sequence has \(d=6\).
    Substituting \(n=1\) into the expression shows that \(t_1=6(1)-2=4\) so that \(a=4\).
  4. \(t_n=\dfrac3n\)is not linear so it is not the general term of an arithmetic sequence.
  5. \(t_n=-4\) is linear so it is the general term of an arithmetic sequence.
    The sequence has \(a=-4\) and \(d=0\); in other words, it is the sequence \(-4,-4,-4,-4, \ldots\)

Example 6

Determine the \(50\)th term of the arithmetic sequence \(-273,-260,-247,-234,\ldots\)

Solution

The sequence is arithmetic with \(a=-273\) and \(d=13\) (by inspection, since, for instance, \(-260-(-273)=13\)).

Thus, the general term is:

\(\begin{align*} t_n &= a+(n-1)d\\ & = -273+(n-1)(13)\\ &= -273 +13(n-1)\\ &= -273+13n-13\\ &= 13n-286 \end{align*}\)

Therefore, we can determine \(t_{50}\): 

\(\begin{align*} t_n &= 13n-286\\ t_{50} &= 13 (50)-286\\ &= 364 \end{align*}\)


Check Your Understanding 5


Determine the simplified general term of the arithmetic sequence \(((((t)*(e))*(r))*(m))*1.0,~ ((((t)*(e))*(r))*(m))*2.0,~ ((((t)*(e))*(r))*(m))*3.0,~ ((((t)*(e))*(r))*(m))*4.0, \ldots\).

The simplified general term, \(t_n\) =   There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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