6 ; 6 ; 9 ; 15 ; .. - NSC Mathematics - Question 3 - 2017 - Paper 1

To derive the general term $T_n$, we use the information from the differences:

1. A quadratic can be expressed as: $T_n = an^2 + bn + c$

2. The known terms give us a system of equations:

   • For $n=1$: $a(1^2) + b(1) + c = 6$
   • For $n=2$: $a(2^2) + b(2) + c = 6$
   • For $n=3$: $a(3^2) + b(3) + c = 9$
   • For $n=4$: $a(4^2) + b(4) + c = 15$

3. Solving these equations, we find:

   • $a = \frac{3}{2}$, $b = -\frac{9}{2}$, $c = 9$.

Thus, the general term is: $T_n = \frac{3}{2}n^2 - \frac{9}{2}n + 9$

To find which term is equal to 3 249, we set the general term equal to 3 249:

1. Equate and simplify: $\frac{3}{2}n^2 - \frac{9}{2}n + 9 = 3249$

2. Rearranging gives: $3n^2 - 9n - 6480 = 0$

3. Using the quadratic formula, $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = -9$, and $c = -6480$, we calculate:

   • $n = \frac{9 \pm \sqrt{(-9)^2 - 4(3)(-6480)}}{2(3)}$
   • Calculate the discriminant and solve for $n$. This resolves to $n = 45$ or $n = -48$, hence:

The term corresponding to a value of 3 249 is the 45th term.

For the sequence to be arithmetic, the difference between consecutive terms must be constant. Calculate the common difference:

1. From the terms:

   • The difference between the second term and the first: $2sin3x - (-1) = 2sin3x + 1$
   • The difference between the third term and the second: $5 - 2sin3x$

2. Set them equal for an arithmetic sequence: $2sin3x + 1 = 5 - 2sin3x$

3. Solve for $x$: $4sin3x = 4 \implies sin3x = 1$

   • This implies: $3x = 90^\circ \implies x = 30^\circ$

Thus, the value of $x$ in the given interval where the sequence is arithmetic is $30$.