1. **Problem 1:** Find the first term $a$ of an AP where the 43rd term $T_{43} = 26$ and the common difference $d = \frac{1}{2}$. 2. The formula for the $n$th term of an AP is $$T_n = a + (n-1)d$$ 3. Substitute $n=43$, $T_{43} = 26$, and $d = \frac{1}{2}$: $$26 = a + (43-1) \times \frac{1}{2} = a + 42 \times \frac{1}{2} = a + 21$$ 4. Solve for $a$: $$a = 26 - 21 = 5$$ --- 5. **Problem 2:** Two APs have the same first and last terms. The first AP has 21 terms with common difference $d_1 = 9$. The second AP has common difference $d_2 = 4$. Find the number of terms $n_2$ in the second AP. 6. Let the first term be $a$ and the last term be $l$. For the first AP: $$l = a + (21 - 1) \times 9 = a + 20 \times 9 = a + 180$$ 7. For the second AP with $n_2$ terms: $$l = a + (n_2 - 1) \times 4$$ 8. Since both APs have the same last term $l$, set equal: $$a + 180 = a + (n_2 - 1) \times 4$$ 9. Simplify and solve for $n_2$: $$180 = (n_2 - 1) \times 4$$ $$n_2 - 1 = \frac{180}{4} = 45$$ $$n_2 = 46$$ **Final answers:** - First term $a = 5$ - Number of terms in second AP $n_2 = 46$