Example 4
Completion requirements
Variables appear in the general term of an arithmetic sequence. They can also appear in the terms of a sequence, as you will investigate in Example 4.
The first three terms of an arithmetic sequence are \(5z + 1 \), \(11z – 5 \), and \(8z + 7 \). Determine the value of \(z \), and state the three terms in the sequence.
The key to this question is understanding how to find the common difference.
Recall that \(d = t_n - t_{n - 1} \). In this case, you subtract the first term from the second term, and the second term from the third term. These two values must be equal because the differences between pairs of consecutive terms in an arithmetic sequence must be constant.
Note that the last line is not \(d \).
This is the value of \(z \).
Using \(z = 2 \), you can determine the three terms.
\(5(2) + 1 = 11 \)
\(11(2) - 5 = 17 \)
\(8(2) + 7 = 23 \)
The three terms are \(11\), \(17\), and \(23\).
Recall that \(d = t_n - t_{n - 1} \). In this case, you subtract the first term from the second term, and the second term from the third term. These two values must be equal because the differences between pairs of consecutive terms in an arithmetic sequence must be constant.
\(
\begin{align}
d &= d \\
t_2 - t_1 &= t_3 - t_2 \\
\left( {11z - 5} \right) - \left( {5z + 1} \right) &= \left( {8z + 7} \right) - \left( {11z - 5} \right) \\
11z - 5 - 5z - 1 &= 8z + 7 - 11z + 5 \\
6z - 6 &= -3z + 12 \\
9z &= 18 \\
z &= 2 \\
\end{align} \)
d &= d \\
t_2 - t_1 &= t_3 - t_2 \\
\left( {11z - 5} \right) - \left( {5z + 1} \right) &= \left( {8z + 7} \right) - \left( {11z - 5} \right) \\
11z - 5 - 5z - 1 &= 8z + 7 - 11z + 5 \\
6z - 6 &= -3z + 12 \\
9z &= 18 \\
z &= 2 \\
\end{align} \)
Note that the last line is not \(d \).
This is the value of \(z \).
Using \(z = 2 \), you can determine the three terms.
\(5(2) + 1 = 11 \)
\(11(2) - 5 = 17 \)
\(8(2) + 7 = 23 \)
The three terms are \(11\), \(17\), and \(23\).