2.1 7 ; x ; y ; -11 ; .. - NSC Mathematics - Question 2 - 2020 - Paper 1

First, observe the pattern:

• The values are: -3, 6, 27, 60.

To find the coefficients:

• Calculate the first differences: $6 - (-3) = 9, \quad 27 - 6 = 21, \quad 60 - 27 = 33$
• Next, the second differences: $21 - 9 = 12, \quad 33 - 21 = 12$

Since the second difference is constant, it suggests a quadratic function. Thus:

1. Set up the equations: $2a = 12 \Rightarrow a = 6$

2. Then use:

\Rightarrow 3(6) + b = 9\n \Rightarrow b = -9$$ 3. Finally, substitute for $c$: $$-3 = 6 + (-9) + c\n \Rightarrow c = 0$$ Thus, the general term is: $$T_n = 6n^2 - 9n$$

To find the 50th term, substitute $n = 50$ into the general term:

= 6(2500) - 450\ = 15000 - 450\ = 14550$$ Thus, the value of the 50th term is $T_{50} = 14550$.

The first differences were found to be: $S_n = 6n^2 + 3n$

To verify:

1. Establish the equations for sum of the first n terms derived from the earlier established differences.
2. Substitute $n$ into the formula, $S_n = n/2 (first term + last term)$ This confirms that this equation holds for any value of $n$.

Let $T_n = -3 + S_n = 21060$.

1. Set up the equation: $6n^2 + 3n - 21063 = 0$
2. Use the quadratic formula, where:

= \frac{-3 \pm \sqrt{(3)^2 - 4(6)(-21063)}}{2(6)}\ = \frac{-3 \pm \sqrt{9 + 504756}}{12}\ = \frac{-3 \pm \sqrt{504765}}{12}\ = \frac{-3 + 711}{12}$$ 3. Thus, $$n = 59$$ So, 59 consecutive first-differences were added.