1. **Problem:** The third term of an arithmetic progression (AP) is 9, and the seventh term is 49. Calculate the thirteenth term. 2. **Formula:** The $n$th term of an AP is given by $$a_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. 3. **Step 1:** Write equations for the given terms: $$a_3 = a + 2d = 9$$ $$a_7 = a + 6d = 49$$ 4. **Step 2:** Subtract the first equation from the second to find $d$: $$a + 6d = 49$$ $$-(a + 2d = 9)$$ \begin{aligned} &\cancel{a} + 6d - \cancel{a} - 2d = 49 - 9 \\ &4d = 40 \\ &d = \frac{40}{4} = 10 \end{aligned}$$ 5. **Step 3:** Substitute $d=10$ into $a + 2d = 9$ to find $a$: $$a + 2(10) = 9$$ $$a + 20 = 9$$ $$a = 9 - 20 = -11$$ 6. **Step 4:** Find the thirteenth term: $$a_{13} = a + 12d = -11 + 12(10) = -11 + 120 = 109$$ **Final answer:** The thirteenth term is $109$.