(B) 16th
Here, $a = -29$, $d = 3$, $a_n = 16$.
Using $a_n = a + (n-1)d$: $16 = -29 + (n-1) \times 3 \Rightarrow 45 = 3(n-1) \Rightarrow n = 16$.
Source: Chapter 5, Section 5.3 (nth term formula)
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Use the formula $a_n = a + (n-1)d$. Here $a = -29$, $d = -26-(-29) = 3$, and $a_n = 16$. Solving gives $n = 16$. A common mistake is arithmetic error when dividing; always verify by substituting $n$ back.