Answer
The common ratio is $\frac{t}{2}$.
The fifth term is $\frac{t^{5}}{16}$.
$a_{n} = t({\frac{t}{2}})^{n-1}$
Work Step by Step
The terms given are t $\frac{t^{2}}{2}$ $\frac{t^{3}}{4}$ $\frac{t^{4}}{8}$.
Since each term is multiplied by $\frac{t}{2}$ to get to the next, the common ratio is $\frac{t}{2}$ The fifth term = the fourth term $\times$ common ratio = $\frac{t^{4}}{8}$ $\times $ $\frac{t}{2}$$ = \frac{t^{5}}{16}$. The formula for the nth term of a geometric sequence is written in the form $a_{n} = a(r)^{n-1}$ since the first term (a) is t and the common ratio (r) is $\frac{t}{2}$.
Plug these into the formula to get $a_{n} = t({\frac{t}{2}})^{n-1}$
