1. **State the problem:** We are given the values $A=129$, $a=4$, and $t=18$ and need to find $T$ in the formula for the sum of an arithmetic series. 2. **Recall the formula:** The sum of the first $n$ terms of an arithmetic sequence is given by $$A = \frac{n}{2} (a + T)$$ where $A$ is the sum, $a$ is the first term, $T$ is the last term, and $n$ is the number of terms. 3. **Identify variables:** Here, $A=129$, $a=4$, and $t=18$ (which represents $n$, the number of terms). 4. **Substitute known values:** $$129 = \frac{18}{2} (4 + T)$$ 5. **Simplify the fraction:** $$129 = 9 (4 + T)$$ 6. **Divide both sides by 9:** $$\frac{129}{9} = 4 + T$$ 7. **Simplify the fraction with cancellation:** $$\frac{\cancel{129}}{\cancel{9}} = 4 + T \quad \Rightarrow \quad 14.3333 = 4 + T$$ 8. **Solve for $T$:** $$T = 14.3333 - 4 = 10.3333$$ **Final answer:** $$\boxed{T = \frac{31}{3} \approx 10.33}$$