Example 2
Use the arithmetic sequence below to answer the following questions.

\(136, 130, 124, 118, 112, …, 4 \)

  1. Determine \(t_1\), \(d\), and the general term, \(t_n\).

    \( \begin{align}
     t_1&= 136 \\ \end{align} \)

    \( \begin{align}
    d&= 130 - 136 = -6 \\  \end{align} \)

    \( \begin{align}
     t_n&= t_1  + \left( {n - 1} \right)d \\
     t_n&= 136 + \left( {n - 1} \right)\left( {-6} \right) \\
     t_n&= 136 - 6n + 6 \\
     t_n&= 142 - 6n \\
     \end{align} \)
  2. What is the \(20^{\rm{th}}\) term?

    Using the general term found above, substitute \(n = 20\), and solve.

    \( \begin{align}
     t_n&= 142 - 6n \\
     t_{20}&= 142 - 6\left( {20} \right) \\
     t_{20}&= 142 - 120 \\
     t_{20}&= 22 \\
     \end{align} \)

    Note that you can write out the entire sequence to determine the 20th term, but using the general term is more efficient, especially when the value of \(n \) is large.

  3. How many terms are in the sequence?

    Using the formula from part a., substitute \(4\) for \(t_n \), and solve for \(n \).

    \( \begin{align}
    t_n&= 142 - 6n \\
    4&= 142 - 6n \\
    -138&= -6n \\
    23&= n \\
     \end{align} \)

    There are \(23\) terms in the given finite arithmetic sequence.