1. **Stating the problem:** We are given a quadratic number pattern: -5, -4, -1, 4, ... We need to find the two consecutive terms in the first differences sequence whose product is 1155. 2. **Find the first differences:** The terms are $T_1 = -5$, $T_2 = -4$, $T_3 = -1$, $T_4 = 4$. Calculate the first differences $d_n = T_{n+1} - T_n$: $$d_1 = T_2 - T_1 = -4 - (-5) = 1$$ $$d_2 = T_3 - T_2 = -1 - (-4) = 3$$ $$d_3 = T_4 - T_3 = 4 - (-1) = 5$$ 3. **Find the second differences:** Since the pattern is quadratic, the second differences are constant: $$d_2 - d_1 = 3 - 1 = 2$$ $$d_3 - d_2 = 5 - 3 = 2$$ So the second difference $= 2$. 4. **General form of first differences:** The first differences form an arithmetic sequence with first term $d_1 = 1$ and common difference $2$. So, $$d_n = d_1 + (n-1) \times 2 = 1 + 2(n-1) = 2n - 1$$ 5. **Find two consecutive first differences whose product is 1155:** We want $d_n \times d_{n+1} = 1155$. Substitute: $$ (2n - 1)(2(n+1) - 1) = 1155 $$ $$ (2n - 1)(2n + 2 - 1) = 1155 $$ $$ (2n - 1)(2n + 1) = 1155 $$ 6. **Expand and solve the quadratic equation:** $$ (2n - 1)(2n + 1) = 4n^2 - 1 = 1155 $$ $$ 4n^2 - 1 = 1155 $$ $$ 4n^2 = 1156 $$ $$ n^2 = 289 $$ $$ n = \pm 17 $$ Since $n$ is a term index, it must be positive, so $n = 17$. 7. **Find the two consecutive first differences:** $$ d_{17} = 2(17) - 1 = 34 - 1 = 33 $$ $$ d_{18} = 2(18) - 1 = 36 - 1 = 35 $$ Check product: $$ 33 \times 35 = 1155 $$ **Final answer:** The two consecutive terms in the first differences sequence are the 17th and 18th terms, which are 33 and 35 respectively.