1. The problem asks to determine the term $T_n$ of the pattern 4, 9, $x$, 37. 2. First, observe the given terms: $T_1 = 4$, $T_2 = 9$, $T_3 = x$, and $T_4 = 37$. 3. We need to find a formula or rule that fits these terms. 4. Check the differences between terms: $T_2 - T_1 = 9 - 4 = 5$ $T_4 - T_2 = 37 - 9 = 28$ 5. Since the difference between $T_2$ and $T_1$ is 5, and between $T_4$ and $T_2$ is 28, the pattern might not be arithmetic. 6. Check if the pattern is quadratic by examining second differences: We don't know $T_3$, so let $T_3 = x$. First differences: $d_1 = 9 - 4 = 5$, $d_2 = x - 9$, $d_3 = 37 - x$. Second differences: $d_2 - d_1 = (x - 9) - 5 = x - 14$, $d_3 - d_2 = (37 - x) - (x - 9) = 46 - 2x$. 7. For a quadratic sequence, second differences are constant, so set: $$x - 14 = 46 - 2x$$ 8. Solve for $x$: $$x - 14 = 46 - 2x$$ $$x + 2x = 46 + 14$$ $$3x = 60$$ $$x = 20$$ 9. So, $T_3 = 20$. 10. Now find the quadratic formula for $T_n = an^2 + bn + c$ using $T_1=4$, $T_2=9$, $T_3=20$: From $n=1$: $a + b + c = 4$ From $n=2$: $4a + 2b + c = 9$ From $n=3$: $9a + 3b + c = 20$ 11. Subtract first from second: $(4a + 2b + c) - (a + b + c) = 9 - 4$ $$3a + b = 5$$ 12. Subtract second from third: $(9a + 3b + c) - (4a + 2b + c) = 20 - 9$ $$5a + b = 11$$ 13. Subtract equation 11 from 12: $$(5a + b) - (3a + b) = 11 - 5$$ $$2a = 6$$ $$a = 3$$ 14. Substitute $a=3$ into equation 11: $$3(3) + b = 5$$ $$9 + b = 5$$ $$b = -4$$ 15. Substitute $a=3$, $b=-4$ into $a + b + c = 4$: $$3 - 4 + c = 4$$ $$-1 + c = 4$$ $$c = 5$$ 16. Therefore, the formula for the $n$th term is: $$T_n = 3n^2 - 4n + 5$$ 17. Verify for $n=4$: $$T_4 = 3(4)^2 - 4(4) + 5 = 3(16) - 16 + 5 = 48 - 16 + 5 = 37$$ This matches the given term. Final answer: $$T_n = 3n^2 - 4n + 5$$