Answer
$a_3=2.56$
$a_5=1.6384$
Work Step by Step
If the original amount of water was $5$ gal and each time they replace $1$ gal (i.e. $20\%$) with antifreeze, then the amount of water decreases each time by $20\%$. Thus, the amount can be modeled by:
$a_n=5(0.8)^{n}$.
Plugging in $n=3$, we get:
$a_3=5(0.8)^{3}=2.56$
Plugging in $n=5$, we get:
$a_5=5(0.8)^{5}=1.6384$
