It looks like the recursive sequence is
[tex]a_n = \dfrac{a_{n-1}}2[/tex]
By substitution, we have
[tex]a_{n-1} =\dfrac{a_{n-2}}2 \implies a_n = \dfrac{a_{n-2}}{2^2}[/tex]
[tex]a_{n-2} = \dfrac{a_{n-3}}2 \implies a_n = \dfrac{a_{n-3}}{2^3}[/tex]
[tex]a_{n-3} = \dfrac{a_{n-4}}2 \implies a_n = \dfrac{a_{n-4}}{2^4}[/tex]
and so on, down to
[tex]a_n = \dfrac{a_1}{2^{n-1}}}[/tex]
Given that [tex]a_1=64[/tex], it follows that the 7th term in the sequence is
[tex]a_7 = \dfrac{64}{2^{7-1}} = \dfrac{64}{2^6} = \dfrac{64}{64} = \boxed{1}[/tex]
