1. Problem: The sum of the first 10 terms ($S_{10}$) of an A.P is 240 and the 8th term ($a_8$) is 34. Find: a. The common difference ($d$) b. The first term ($a$) c. The sum of the first 30 terms ($S_{30}$) Step 1: Let the first term be $a$ and the common difference be $d$. Step 2: Use the nth term formula of A.P: $$a_n = a + (n-1)d$$ Given $a_8 = 34$: $$a + 7d = 34 \, \Rightarrow (1)$$ Step 3: Use the sum of n terms formula of A.P: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ Given $S_{10} = 240$: $$240 = \frac{10}{2}[2a + 9d] \, \Rightarrow 240 = 5(2a + 9d) \Rightarrow 48 = 2a + 9d \, \Rightarrow (2)$$ Step 4: Solve equations (1) and (2): From (1): $$a = 34 - 7d$$ Plug into (2): $$48 = 2(34 - 7d) + 9d = 68 - 14d + 9d = 68 - 5d$$ $$48 - 68 = -5d \Rightarrow -20 = -5d \Rightarrow d = 4$$ Step 5: Find $a$: $$a = 34 - 7(4) = 34 - 28 = 6$$ Step 6: Find $S_{30}$: $$S_{30} = \frac{30}{2}[2(6) + 29(4)] = 15[12 + 116] = 15 \times 128 = 1920$$ --- 2. Problem: The first three terms of an A.P are $2x$, $3x - 1$, and $x + 4$. Find: i. The value of $x$ ii. The sequence iii. The 19th term Step 1: For A.P, the difference between consecutive terms is constant: $$ (3x - 1) - 2x = (x + 4) - (3x - 1)$$ Simplify: $$x - 1 = x + 4 - 3x + 1 = -2x + 5$$ $$x -1 = -2x + 5$$ $$x + 2x = 5 + 1$$ $$3x = 6 \Rightarrow x = 2$$ Step 2: Substitute $x=2$ into terms: $$a_1 = 2x = 4$$ $$a_2 = 3x -1 = 6 -1 = 5$$ $$a_3 = x + 4 = 6$$ Step 3: Common difference: $$d = a_2 - a_1 = 5 - 4 = 1$$ Step 4: The sequence is: $$4, 5, 6, 7, ...$$ Step 5: Find the 19th term: $$a_{19} = a_1 + 18d = 4 + 18(1) = 22$$ --- 3. Problem: In an A.P, the 4th term is 20 and the 8th term is 1.5 times the 4th term. Find: a. The common difference b. The first term c. The sequence Step 1: Let $a$ be the first term and $d$ the common difference. Step 2: Use nth term formula: $$a_4 = a + 3d = 20 \quad (1)$$ $$a_8 = a + 7d = 1.5 \times 20 = 30 \quad (2)$$ Step 3: Subtract (1) from (2): $$a + 7d - (a + 3d) = 30 - 20$$ $$4d = 10 \Rightarrow d = \frac{10}{4} = 2.5$$ Step 4: Substitute $d = 2.5$ into (1): $$a + 3(2.5) = 20 \Rightarrow a + 7.5 = 20 \Rightarrow a = 12.5$$ Step 5: The sequence is: $$a_n = 12.5 + (n-1)2.5$$ Examples: $12.5, 15, 17.5, 20, 22.5, ...$ --- 4. Problem: Evaluate $$\sum_{r=1}^{18} (3r + 2)$$ Step 1: Split the summation: $$\sum_{r=1}^{18} (3r + 2) = 3 \sum_{r=1}^{18} r + 2 \sum_{r=1}^{18} 1$$ Step 2: Use formula for sum of first $n$ natural numbers: $$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$ Step 3: Calculate: $$3 \times \frac{18 \times 19}{2} + 2 \times 18 = 3 \times 171 + 36 = 513 + 36 = 549$$