Here, $a = -\dfrac{1}{3}$, $d = \dfrac{2}{3} - \left(-\dfrac{1}{3}\right) = 1$.
$a_n = a + (n-1)d = -\dfrac{1}{3} + (n-1)(1) = n - 1 - \dfrac{1}{3} = n - \dfrac{4}{3}$
Answer: (B) $n - \dfrac{4}{3}$
Use $a_n = a + (n-1)d$ with $a = -\frac{1}{3}$ and $d = 1$. Simplify carefully: $-\frac{1}{3} + n - 1 = n - \frac{4}{3}$. The trap options mix up signs or place the fraction incorrectly — always simplify step by step. Source: Chapter 5, Section 5.3 (nth term formula).