Answer
$r=\dfrac{t}{2}$
$a_5=\dfrac{t^5}{16}$
The $n^{th}$ term of the geometric sequence is: $a_n=t \cdot \left(\dfrac{t}{2}\right)^{n-1}$.
Work Step by Step
RECALL:
(1) The common ratio of a geometric sequence is equal to the quotient of any two consecutive terms:
$r =\dfrac{a_n}{a_{n-1}}$
(2) The $n^{th}$ term of a geometric sequence is given by the formula:
$a_n = a\cdot r^{n-1}$
where
$a$ = first term
$r$ = common ratio
The sequence is said to be geometric.
Thus, we can proceed to solving for the common ratio:
$\require{cancel}
r=\dfrac{\frac{t^2}{2}}{t}
\\r=\dfrac{t^2}{2} \cdot \dfrac{1}{t}
\\r=\dfrac{\cancel{t^2}t}{2} \cdot \dfrac{1}{\cancel{t}}
\\r=\dfrac{t}{2}$
The fifth term can be found by multiplying the common ratio to the fourth term.
The fourth term is $\dfrac{t^4}{8}$.
Thus, the fifth term is:
$a_5=\dfrac{t^4}{8} \cdot \dfrac{t}{2}$
Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain:
$a_5=\dfrac{t^{4+1}}{16}
\\a_5=\dfrac{t^5}{16}$
With a first term of $t$ and a common ratio of $r=\dfrac{t}{2}$, the $n^{th}$ term of the geometric sequence is:
$a_n=a \cdot r^{n-1}
\\a_n=t \cdot \left(\dfrac{t}{2}\right)^{n-1}$
