1. **Calculate the area of quadrilateral ABCD** Since the problem does not provide coordinates or side lengths, we cannot calculate the area without additional information. --- 2. **Complete the table for sequences A, B, C, and D** - Sequence A: 2, 5, 8, 11 - This is an arithmetic sequence with common difference $3$. - $n$th term formula: $a_n = 2 + (n-1) \times 3 = 3n - 1$ - Next term: $11 + 3 = 14$ - Sequence B: 20, 14, 8, 2 - Arithmetic sequence with common difference $-6$. - $n$th term: $a_n = 20 + (n-1)(-6) = 26 - 6n$ - Next term: $2 - 6 = -4$ - Sequence C: 1, 4, 9, 16 - These are perfect squares: $1^2, 2^2, 3^2, 4^2$ - $n$th term: $a_n = n^2$ - Next term: $5^2 = 25$ - Sequence D: 0, 2, 6, 12 - Differences: 2, 4, 6 (increasing by 2) - $n$th term formula: $a_n = n(n-1)$ (product of consecutive integers) - Next term: $5 imes 4 = 20$ --- 3. **Sum of first $n$ terms is $\frac{n(3n+1)}{2}$** (i) When sum is 155, set equation: $$\frac{n(3n+1)}{2} = 155$$ Multiply both sides by 2: $$n(3n+1) = 310$$ Expand: $$3n^2 + n = 310$$ Rearranged: $$3n^2 + n - 310 = 0$$ (ii) Solve quadratic $3n^2 + n - 310 = 0$ using quadratic formula: $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=3$, $b=1$, $c=-310$. Calculate discriminant: $$\Delta = 1^2 - 4 \times 3 \times (-310) = 1 + 3720 = 3721$$ Square root: $$\sqrt{3721} = 61$$ Solutions: $$n = \frac{-1 \pm 61}{6}$$ - Positive root: $$n = \frac{-1 + 61}{6} = \frac{60}{6} = 10$$ - Negative root: $$n = \frac{-1 - 61}{6} = \frac{-62}{6} = -\frac{31}{3}$$ (discarded since $n$ must be positive integer) (iii) Complete statement: The sum of the first **10** terms of this sequence is 155. --- 4. **Find $w$ given area of sector is $r^2$ cm² and radius $r$ cm** Formula for area of sector: $$\text{Area} = \frac{w}{360} \times \pi r^2$$ Given: $$\frac{w}{360} \pi r^2 = r^2$$ Divide both sides by $r^2$ (assuming $r \neq 0$): $$\frac{w}{360} \pi = 1$$ Solve for $w$: $$w = \frac{360}{\pi}$$ Numerical approximation: $$w \approx \frac{360}{3.1416} \approx 114.59$$ **Final answer:** $$w = \frac{360}{\pi}$$ degrees