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Determine \(t_1\), \(d\), and the missing terms of the following sequence:

\(\_\_\_,15,\_\_\_,\_\_\_,72\)
    Equation I

    For this question, you can set up a system of linear equations, similar to Example 5.

    \( \begin{align}
     t_2&= 15 \\
     n &= 2 \\
     d &= ? \\
     t_1 &= ? \\
     \end{align} \)
    \( \begin{align}
     t_n &= t_1 + \left( {n - 1} \right)d \\
     15 &= t_1 + \left( {2 - 1} \right)d \\
     15 &= t_1 + d \\
     \end{align} \)


    Equation II

    \( \begin{align}
     t_5 &= 72 \\
     n &= 5 \\
     d &= ? \\
     t_1 &= ? \\
     \end{align} \)
    \(\begin{align}
     t_n &= t_1 + \left( {n - 1} \right)d \\
     72 &= t_1 + \left( {5 - 1} \right)d \\
     72 &= t_1 + 4d \\
     \end{align}\)


    Subtract Equation I from Equation II

    \[\begin{align}
     72 &= \cancel{t_1}  + 4d \\
    15 &= \cancel{t_1} + d \\
    \hline \\
     57 &= \qquad 3d \\
     19 &= \qquad \enspace d \\
     \end{align}\]


    Then find \(t_1 \).

    \(\begin{align}
     15 &= t_1 + d \\
     15 &= t_1 + 19 \\
     âˆ’4 &= t_1  \\
     \end{align} \)

    To find the two missing terms, add \(19\) to the previous term.

    \(\begin{align}
     t_3 &= 15 + 19 = 34 \\
     t_4 &= 34 + 19 = 53 \\
     \end{align} \)

    The missing terms are  \(−4, 15, 34, 53, 72\).

    Alternatively, you can say that the two terms are three steps from each other; therefore, \(3d = (72 − 15) \) or \(3d = 57 \).