Arithmetic Sequences
An arithmetic sequence is a sequence where the difference between consecutive terms is a constant. In other words, you are either adding or subtracting the same number to get your next term. An example of an arithmetic sequence would be \(7, 4, 1, −2, −5,...\) because the difference between consecutive terms is \(−3\). Another example would be \(5, \frac{11}{2}, 6, \frac{13}{2}, 7, \frac{15}{2},...\) because the difference between consecutive terms is \(\frac{1}{2}\). If we thought of these sequences as a set of ordered pairs with the domain being the term number (consecutive integers starting at \(1\) and the range being the terms of the sequence and then plotted them (see below), you will notice that these points would lie on the same line and follow a linear pattern. The slope of this line formed by the points is the common difference. We will use \(d\) to notate the common difference.
An arithmetic sequence is a sequence where the difference between consecutive terms is a constant. In other words, you are either adding or subtracting the same number to get your next term. An example of an arithmetic sequence would be \(7, 4, 1, −2, −5,...\) because the difference between consecutive terms is \(−3\). Another example would be \(5, \frac{11}{2}, 6, \frac{13}{2}, 7, \frac{15}{2},...\) because the difference between consecutive terms is \(\frac{1}{2}\). If we thought of these sequences as a set of ordered pairs with the domain being the term number (consecutive integers starting at \(1\) and the range being the terms of the sequence and then plotted them (see below), you will notice that these points would lie on the same line and follow a linear pattern. The slope of this line formed by the points is the common difference. We will use \(d\) to notate the common difference.
Because of the linear relationship that arithmetic sequences have, we can use point-slope form to help us derive the explicit formula for arithmetic sequences. We will use the first term of a sequence as a point \((1,a_1)\) and the common difference \(d\) as the slope.
\(\begin{align}&y−y_1=m(x−x_1)\\
&y−a_1=d(x−1)\ & \ \ \ \ &\text{1) Substitute the point and slope described above.}\\
&a_n−a_1=d(x−1)\ & \ \ \ \ &\text{2) Let x (the domain) be the term number and y( the range) be the term}\ a_n.\\
&a_n=a_1+d(n−1)\ & \ \ \ \ &\text{3) Solve for}\ a_n.\end{align}\)
Explicit Formula for an Arithmetic Sequence: \(a_n=a_1+d(n−1)\)
Example 1: We’ll first go back to one of the sequences we talked about above: \(5, \frac{11}{2}, 6, \frac{13}{2}, 7, \frac{15}{2},...\). We know right away what our two key pieces of information are. The first term \(a_1\) is \(5\), and the common difference \(d\) is \(\frac{1}{2}\). The common difference can always be found by finding the difference between a term and its previous term (we’ll look later at what to do if you don’t have consecutive terms). So we can substitute these into the explicit formula.
\(\begin{align}&a_n=5+\frac{1}{2}(n−1)\ & \ \ \ &\text{1) Substitute}\ a_1 \text{and d into the explicit formula.}\\
&a_n=5+\frac{1}{2}n−\frac{1}{2}\ & \ \ \ &\text{2) Distribute.}\\
&a_n=\frac{1}{2}n+\frac{9}{2}\ & \ \ \ &\text{3) Combine like terms.}\end{align}\)
We can check our work using any of the other terms we know. For example, we know that the fourth term should be \(\frac{13}{2}\). If we substitute \(4\) in for \(n\) into our formula, we do get \(\frac{13}{2}\).
Example 2: Find the \(34th\) term of the arithmetic sequence that has a common difference of \(-4\) and the seventh term is \(92\).
In order to find the \(34th\) term, we first will find the explicit formula. We know that \(d=−4\) and that \(a_7=92\). So we can substitute \(-4\) in for \(d\), \(7\) in for \(n\), and \(92\) in for \(a_n\) which is really \(a_7\) since we are letting \(n=7\).
\(\begin{align}&a_n=a_1+d(n−1)\\\\
&92=a_1−4(7−1)\ & \ \ &\text{1) Substitute values in for d, n and}\ a_n.\\\\
&92=a_1−4(6) \ & \ \ &\text{2) Solve for}\ a_1.\\\\
&92=a_1−24\\\\
&a_1=116\ & \ \ &\text{This is the first term of the sequence so we can now find the explicit formula.}\\\\
&a_n=116−4(n−1)\ & \ \ &\text{3) Substitute -4 in for d and 116 for}\ a_1.\\\\
&a_n=116−4n+4\ & \ \ &\text{4) Simplify.}\\\\
&a_n=−4n+120\ & \ \ &\text{This is the explicit formula.}\end{align}\)
We will use the explicit formula to now find the \(34th\) term.
\(a_34=−4(34)+120\)
\(a_34=−16\)
The \(34th\) term in this sequence is \(-16\).
Example 3: The tenth term of an arithmetic sequence is \(-65\) and the twenty-eighth term is \(-191\). Write the explicit formula for the sequence. Then find the value of the \(45th\) term.