1. **Arithmetic Progressions (AP) Problem:** State the problem: Find the 10th term and the sum of the first 10 terms of an AP where the first term $a_1=3$ and common difference $d=5$. Formula used: - $n$th term: $a_n = a_1 + (n-1)d$ - Sum of first $n$ terms: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ Step-by-step: 1. Calculate the 10th term: $a_{10} = 3 + (10-1) \times 5 = 3 + 45 = 48$ 2. Calculate the sum of first 10 terms: $S_{10} = \frac{10}{2}(2 \times 3 + 9 \times 5) = 5(6 + 45) = 5 \times 51 = 255$ 2. **Simultaneous Equations Problem:** State the problem: Solve the system: $$\begin{cases} 2x + 3y = 12 \\ 5x - y = 13 \end{cases}$$ Method: Use substitution or elimination. Step-by-step: 1. From second equation: $5x - y = 13 \Rightarrow y = 5x - 13$ 2. Substitute into first equation: $2x + 3(5x - 13) = 12$ 3. Simplify: $2x + 15x - 39 = 12 \Rightarrow 17x = 51 \Rightarrow x = 3$ 4. Substitute $x=3$ into $y = 5x - 13$: $y = 15 - 13 = 2$ 3. **Sample MCQ Question:** What is the sum of the first 5 terms of the arithmetic progression 2, 6, 10, 14, ...? A) 30 B) 40 C) 50 D) 60 Solution: - First term $a_1=2$, common difference $d=4$ - Sum $S_5 = \frac{5}{2}(2 \times 2 + (5-1) \times 4) = \frac{5}{2}(4 + 16) = \frac{5}{2} \times 20 = 50$ Answer: C) 50