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

Using the standard form of a quadratic sequence, we can express the $n^{th}$ term as:

$T_n = a n^2 + b n + c$

To find the coefficients $a$, $b$, and $c$, we can set up the equations:

1. For $n=1$:
   $T_1 = 6 \Rightarrow a(1)^2 + b(1) + c = 6.$
   (Equation 1)

2. For $n=2: $T_2 = 6 \Rightarrow a(2)^2 + b(2) + c = 6.$
   (Equation 2)

3. For $n=3: $T_3 = 9 \Rightarrow a(3)^2 + b(3) + c = 9.$
   (Equation 3)

Solving these equations will yield:

• From these equations, we can derive: $a = \frac{3}{2},\ b = -\frac{9}{2},\ c = 9.$

Thus, the general term of the pattern is:

$T_n = \frac{3}{2} n^2 - \frac{9}{2} n + 9.$

To find which term equals 3249, we set the general term equal to 3249:

$\frac{3}{2} n^2 - \frac{9}{2} n + 9 = 3249$

Rearranging this gives:

$\frac{3}{2} n^2 - \frac{9}{2} n - 3240 = 0.$

Multiplying by 2 to eliminate the fraction results in:

$3n^2 - 9n - 6480 = 0.$

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

Calculating the discriminant: $b^2 - 4ac = (-9)^2 - 4(3)(-6480) = 81 + 77760 = 77741.$

Substituting values into the quadratic formula:

$n = \frac{9 \pm \sqrt{77741}}{6}.$

Calculating $n$, we find possible values for n, leading to the final result:

• $n = 45$ or $n = -48$.

Therefore, the term which has a value of 3249 is the 45th term.

For an arithmetic sequence, the difference between consecutive terms must be constant. Thus:

The common difference: $2\sin(3x) - (-1) = 5 - 2\sin(3x)$

Setting these equal gives:

$2\sin(3x) + 1 = 5 - 2\sin(3x).$

Rearranging this:

$4\sin(3x) = 4 \Rightarrow \sin(3x) = 1.$

The general solution for $heta$ where $heta = 3x$ is: $3x = 90 + k(360)\ \text{ or } \ 3x = 270 + k(360) \quad (k \text{ is an integer}).$

Dividing by 3: $x = 30 + k(120) \quad (k \text{ is an integer}) \text{ or } \ x = 90 + k(120).$

Only $x = 30$ is valid in the given interval $[0; 90]$.

Thus, the value of $x$ is 30.