= Solution
{c}
Enumerate both <sets> of <peaks of a sequence> in increasing order of index. If $a_n\to L$, the two selected <subsequences> both tend to $L$.
For the converse, suppose $a_{k_r}\to L$ and $a_{m_r}\to L$. Given $\varepsilon>0$, choose $r$ large enough that both selected terms differ from $L$ by less than $\varepsilon$. For every $n>\max\{k_r,m_r\}$, the two <peak of a sequence> conditions imply
$$
\boxed{L-\varepsilon<a_{m_r}\leq a_n\leq a_{k_r}<L+\varepsilon.}
$$
The lower bound comes from the <peak of a sequence> of $(-a_n)$ at $m_r$. This is precisely the definition of $a_n\to L$. Thus \b[convergence is equivalent to a common finite <limit of a sequence> for the upper and lower peak <subsequences>.]
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