10. Given that an infinite arithmetic sequence contains 3 terms (arranged in order but not necessarily consecutive): $13, 25, 41$. Then 2013 $\qquad$ (fill in “is” “is not” or “cannot be determined”) a term in the sequence.
10. Yes.

Let the arithmetic sequence that satisfies the conditions be $\left\{a_{n}\right\}$, with a common difference of $d$.
Without loss of generality, let
$$
a_{1}=13, a_{m}=25, a_{n}=41(n>m>1) \text {. }
$$

Then $d=\frac{25-13}{m-1}=\frac{41-13}{n-1}$.
Thus, $(m-1) d=12,(n-1) d=28$.
Notice that,
$$
\begin{array}{l}
2013=13+2000 \\
=13+(8 \times 12+68 \times 28) \\
=13+8(m-1) d+68(n-1) d \\
=13+(8 m+68 n-76) d .
\end{array}
$$

Therefore, 2013 is the $(8 m+68 n-75)$-th term of the sequence.