= Solution
{c}
It is useful to prove the <contrapositive>. Suppose $k$ is a <peak of a sequence> $(a_n)$ and $m$ is a <peak of a sequence> $(-a_n)$. The second condition says $a_j\geq a_m$ for every $j>m$. Hence
$$
j>\max\{k,m\}\quad\Longrightarrow\quad a_m\leq a_j\leq a_k.
$$
Only finitely many earlier terms remain, and their <absolute values> have a finite <maximum>. Together these give a bound for every term, so $(a_n)$ is a <bounded sequence>. Consequently,
$$
\boxed{(a_n)\text{ unbounded}\ \Longrightarrow\ (a_n)\text{ or }(-a_n)\text{ has no peaks}.}
$$
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