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Determine if the following infinite sequences are geometric. If the sequence is geometric, provide the general term. Justify your decisions.

1. \(1, 5, 25, 125, …\)
   
   
   Solution

   \(\begin{align}
    r &= \frac{{t_n }}{{t_{n - 1} }} \\
    r &= \frac{5}{1} = 5 \\
    r &= \frac{{25}}{5} = 5 \\
    r &= \frac{{125}}{{25}} = 5 \\
    \end{align} \)
   Because \(r = 5 \) and is constant, this is a geometric sequence.
   
   The general term is \(t_n = 1\left( 5 \right)^{n - 1} \) or \(t_n = 5^{n - 1} \).
   
   
   
2. \(15, - 15, 15, - 15, ...\)
   
   
   Solution

   \[\begin{align}
    r &= \frac{{t_n }}{{t_{n - 1} }} \\
    r &= \frac{{ -15}}{{15}} = -1 \\
    r &= \frac{{15}}{{ - 15}} = -1 \\
    r &= \frac{{ -15}}{{15}} = -1 \\
    \end{align}\]

   
   Because \(r = -1 \) and is constant, this is a geometric sequence.
   
   The general term is \(t_n = 15\left( {-1} \right)^{n - 1} \).

   
   Note that the brackets are very important here.  In order for the exponent to apply to both the negative and the \(1\), brackets must be present!

   
   
3. \(16, 4, 1, \frac{1}{4}, ... \)
   
   
   Solution

   \[\begin{align}
    r &= \frac{{t_n }}{{t_{n - 1} }} \\
    r &= \frac{4}{{16}} = \frac{1}{4} \\
    r &= \frac{1}{4} = \frac{1}{4} \\
    r &= \frac{{\frac{1}{4}}}{1} = \frac{1}{4} \\
    \end{align} \]

   
   Because \(r = \frac{1}{4}\) and is constant, this is a geometric sequence.
   
   The general term is \(t_n = 16\left( {\frac{1}{4}} \right)^{n - 1} \).
   
   
   
4. \(2, - 6, 24, - 120, ... \)
   
   
   Solution

   \[\begin{align}
    r &= \frac{{t_n }}{{t_{n - 1} }} \\
    r &= \frac{{ -6}}{2} = -3 \\
    r &= \frac{{24}}{{ - 6}} = -4 \\
    \end{align} \]

   
   Because \(r \) is not constant, this is not a geometric sequence, and a general term cannot be determined.