1. **State the problem:** Given an arithmetic sequence with the last term $a_n = 4$, common difference $d = 2$, and sum of $n$ terms $S_n = -14$, find $n$ (the number of terms) and $a$ (the first term). 2. **Recall formulas:** - The $n$th term of the arithmetic sequence is given by $$a_n = a + (n-1)d$$ - The sum of the first $n$ terms is $$S_n = \frac{n}{2}(a + a_n)$$ 3. **Use the given values:** - Substitute $a_n = 4$ and $d = 2$ into the first formula: $$4 = a + (n-1) \times 2$$ which simplifies to $$a = 4 - 2(n-1) = 4 - 2n + 2 = 6 - 2n$$ 4. **Substitute into the sum formula:** Replace $a$ and $a_n$ in the sum formula: $$-14 = \frac{n}{2}(a + 4) = \frac{n}{2}((6 - 2n) + 4) = \frac{n}{2}(10 - 2n)$$ Simplify the right side: $$-14 = \frac{n}{2} \times (10 - 2n) = \frac{n}{2} \times 2(5 - n) = n(5 - n)$$ Therefore, $$n(5 - n) = -14$$ 5. **Solve the quadratic equation:** Rewrite as: $$5n - n^2 = -14$$ Bring all terms to one side: $$-n^2 + 5n + 14 = 0$$ Multiply both sides by -1 to get a standard form: $$n^2 - 5n - 14 = 0$$ 6. **Apply the quadratic formula:** $$n = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 1 \times (-14)}}{2 \times 1} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2}$$ So, $$n = \frac{5 \pm 9}{2}$$ 7. **Evaluate the roots:** - $$n = \frac{5 + 9}{2} = \frac{14}{2} = 7$$ - $$n = \frac{5 - 9}{2} = \frac{-4}{2} = -2$$ Since $n$ must be positive, reject $n = -2$ and use $n = 7$. 8. **Find $a$:** Using $a = 6 - 2n$, $$a = 6 - 2 \times 7 = 6 - 14 = -8$$ **Final answer:** $$n = 7, \quad a = -8$$