Calculus: Early Transcendentals (2nd Edition)

a) $2$ and $1$ b) $a_{n+1} = \frac{1}{2}a_n$, $a_1 = 64$ c) $a_n = 64\times (\frac{1}{2})^{n-1}$
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. $a_n = 64\times (\frac{1}{2})^{n-1}$