Answer
a) $2$ and $1$
b) $a_{n+1} = \frac{1}{2}a_n$, $a_1 = 64$
c) $a_n = 64\times (\frac{1}{2})^{n-1}$
Work Step by Step
a) As we progress through the sequence, the next term is the previous term halved. Thus, the next two terms are $2$ and $1$.
b) The first term is $64$, so $a_1 = 64$
The next term is the previous term halved:
$a_{n+1} = \frac{1}{2}a_n$
c) The term is $64$ divided by two, $n-1$ times.
$a_n = 64\times (\frac{1}{2})^{n-1}$
