1. **Stating the problem:** We have a sequence with nth term given by $$T_n = an^3 - 2n^2 + n + d$$ where $a$ and $d$ are constants. The sequence has a third difference of 12. The first term is 4. We need to: (i) Find $a$ and $d$. (ii) Write down the next four terms of the sequence. 2. **Recall the formula and important rules:** - The third difference of a cubic sequence is constant and equals $6a$. - The first term is $T_1 = a(1)^3 - 2(1)^2 + 1 + d = a - 2 + 1 + d = a - 1 + d$. 3. **Use the third difference to find $a$:** Given third difference = 12, and third difference for cubic is $6a$, so: $$6a = 12$$ $$a = \frac{12}{6} = 2$$ 4. **Use the first term to find $d$:** Given $T_1 = 4$, substitute $a=2$: $$T_1 = 2 - 1 + d = 1 + d = 4$$ $$d = 4 - 1 = 3$$ 5. **Write the nth term with found constants:** $$T_n = 2n^3 - 2n^2 + n + 3$$ 6. **Calculate the next four terms:** We know $T_1 = 4$. Calculate $T_2, T_3, T_4, T_5$: $$T_2 = 2(2)^3 - 2(2)^2 + 2 + 3 = 2(8) - 2(4) + 2 + 3 = 16 - 8 + 2 + 3 = 13$$ $$T_3 = 2(3)^3 - 2(3)^2 + 3 + 3 = 2(27) - 2(9) + 3 + 3 = 54 - 18 + 6 = 42$$ $$T_4 = 2(4)^3 - 2(4)^2 + 4 + 3 = 2(64) - 2(16) + 4 + 3 = 128 - 32 + 7 = 103$$ $$T_5 = 2(5)^3 - 2(5)^2 + 5 + 3 = 2(125) - 2(25) + 5 + 3 = 250 - 50 + 8 = 208$$ **Final answers:** (i) $a=2$, $d=3$ (ii) Next four terms: 13, 42, 103, 208