1. **Problem statement:** The sum of the first two terms of an arithmetic series is 47, and the thirtieth term is -62. Find: a) The first term and the common difference. b) The sum of the first 60 terms. 2. **Formulas and rules:** - The $n$th term of an arithmetic series is given by: $$a_n = a_1 + (n-1)d$$ where $a_1$ is the first term and $d$ is the common difference. - The sum of the first $n$ terms is: $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 3. **Given:** - $a_1 + a_2 = 47$ - $a_{30} = -62$ 4. **Express $a_2$ in terms of $a_1$ and $d$:** $$a_2 = a_1 + d$$ So, $$a_1 + a_2 = a_1 + (a_1 + d) = 2a_1 + d = 47$$ 5. **Express $a_{30}$ in terms of $a_1$ and $d$:** $$a_{30} = a_1 + 29d = -62$$ 6. **Solve the system:** From step 4: $$2a_1 + d = 47 \implies d = 47 - 2a_1$$ Substitute into step 5: $$a_1 + 29(47 - 2a_1) = -62$$ Simplify: $$a_1 + 1363 - 58a_1 = -62$$ $$-57a_1 = -1425$$ $$a_1 = \frac{1425}{57} = 25$$ 7. **Find $d$:** $$d = 47 - 2(25) = 47 - 50 = -3$$ 8. **Sum of first 60 terms:** Use formula: $$S_{60} = \frac{60}{2} [2(25) + (60 - 1)(-3)] = 30 [50 + 59(-3)] = 30 [50 - 177] = 30 (-127) = -3810$$ **Final answers:** a) First term $a_1 = 25$, common difference $d = -3$. b) Sum of first 60 terms $S_{60} = -3810$.