= Solution
{c}
\b[True.] Choose $N$ such that $\ell/2<a_n<3\ell/2$ for every $n\geq N$. All the finitely many preceding terms are also positive. Define the positive <minimum> and finite <maximum>
$$
\eta=\min\bigl(\{\ell/2\}\cup\{a_j:1\leq j<N\}\bigr)>0,
\qquad
B=\max\bigl(\{3\ell/2\}\cup\{a_j:1\leq j<N\}\bigr)<\infty.
$$
Every term lies between $\eta$ and $B$, so every ratio is at most $B/\eta$. Consequently the required <supremum> satisfies
$$
\boxed{\sup\{a_n/a_m:m,n\in\mathbb N\}\leq B/\eta<\infty.}
$$
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