1. **State the problem:** We have an arithmetic sequence with the first three terms: $8p$, $7p - 3$, and $4p + 2$. The sum of the first $n$ terms is $-1914$. We need to find the value of $n$. 2. **Find the common difference $d$:** The common difference in an arithmetic sequence is the difference between consecutive terms. $$d = (7p - 3) - 8p = 7p - 3 - 8p = -p - 3$$ Check with the next difference: $$(4p + 2) - (7p - 3) = 4p + 2 - 7p + 3 = -3p + 5$$ For the sequence to be arithmetic, these must be equal: $$-p - 3 = -3p + 5$$ Solve for $p$: $$-p - 3 = -3p + 5$$ $$-p + 3p = 5 + 3$$ $$2p = 8$$ $$p = 4$$ 3. **Substitute $p=4$ back into the terms:** First term $a_1 = 8p = 8 \times 4 = 32$ Second term $a_2 = 7p - 3 = 7 \times 4 - 3 = 28 - 3 = 25$ Third term $a_3 = 4p + 2 = 4 \times 4 + 2 = 16 + 2 = 18$ 4. **Find the common difference $d$ with $p=4$:** $$d = a_2 - a_1 = 25 - 32 = -7$$ 5. **Sum of the first $n$ terms formula:** $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ Given $S_n = -1914$, substitute values: $$-1914 = \frac{n}{2} (2 \times 32 + (n-1)(-7))$$ Simplify inside the parentheses: $$2 \times 32 = 64$$ $$-1914 = \frac{n}{2} (64 - 7(n-1))$$ $$-1914 = \frac{n}{2} (64 - 7n + 7)$$ $$-1914 = \frac{n}{2} (71 - 7n)$$ Multiply both sides by 2: $$-3828 = n(71 - 7n)$$ Distribute $n$: $$-3828 = 71n - 7n^2$$ Rearranged: $$7n^2 - 71n - 3828 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$n = \frac{71 \pm \sqrt{(-71)^2 - 4 \times 7 \times (-3828)}}{2 \times 7}$$ Calculate discriminant: $$(-71)^2 = 5041$$ $$4 \times 7 \times 3828 = 107184$$ $$\sqrt{5041 + 107184} = \sqrt{112225} = 335$$ 7. **Calculate $n$ values:** $$n = \frac{71 \pm 335}{14}$$ Two solutions: $$n = \frac{71 + 335}{14} = \frac{406}{14} = 29$$ $$n = \frac{71 - 335}{14} = \frac{-264}{14} = -18.857...$$ Since $n$ must be positive integer, $n = 29$. **Final answer:** $$\boxed{29}$$