1. Problem 1: Given sum of first 10 terms $S_{10} = 240$ and the 8th term $a_8 = 34$ of an arithmetic progression (A.P), find: 1.a Common difference $d$ 1.b First term $a$ 1.c Sum of first 30 terms $S_{30}$ Step 1: Use the formula for the sum of the first $n$ terms of an A.P: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ For $n=10$: $$240 = \frac{10}{2}[2a + 9d] \implies 240 = 5(2a + 9d) \implies 48 = 2a + 9d \ (1)$$ Step 2: Use the formula for the $n$th term of an A.P: $$a_n = a + (n-1)d$$ For $n=8$: $$34 = a + 7d \ (2)$$ Step 3: Solve equations (1) and (2): Multiply (2) by 2: $$68 = 2a + 14d$$ Subtract (1) from this: $$68 - 48 = (2a + 14d) - (2a + 9d) \implies 20 = 5d \implies d = 4$$ Step 4: Substitute $d=4$ into (2): $$34 = a + 7(4) = a + 28 \implies a = 6$$ Step 5: Find the sum of first 30 terms: $$S_{30} = \frac{30}{2}[2(6) + 29(4)] = 15[12 + 116] = 15 \times 128 = 1920$$ 2. Problem 2: First three terms of A.P are $2x$, $3x-1$, and $x+4$. Find: 2.i Value of $x$ Step 1: In an A.P, the difference between consecutive terms is constant: $$(3x-1) - 2x = (x+4) - (3x -1)$$ Simplify: $$x - 1 = x + 4 - 3x + 1 \implies x - 1 = 5 - 2x$$ Add $2x$ to both sides: $$3x - 1 = 5$$ Add 1: $$3x = 6 \implies x = 2$$ 2.ii Sequence Substitute $x=2$: First term: $2x = 4$ Second term: $3x - 1 = 6 - 1 = 5$ Third term: $x + 4 = 6$ Sequence: $4, 5, 6, ...$ 2.iii Find the 19th term: Common difference $d = 5 - 4 = 1$ $$a_{19} = a + 18d = 4 + 18(1) = 22$$ 3. Problem 3: Given $a_4 = 20$ and $a_8 = 1.5 a_4$, find: 3.a Common difference $d$ 3.b First term $a$ 3.c The sequence Step 1: Use formulas: $$a_4 = a + 3d = 20 \ (1)$$ $$a_8 = a + 7d = 1.5 imes 20 = 30 \ (2)$$ Step 2: Subtract (1) from (2): $$(a + 7d) - (a + 3d) = 30 - 20 \implies 4d = 10 \implies d = 2.5$$ Step 3: Substitute $d = 2.5$ into (1): $$a + 3(2.5) = 20 \implies a + 7.5 = 20 \implies a = 12.5$$ Step 4: Write the sequence: $$a_n = 12.5 + (n-1)2.5$$ Sequence starts: $12.5, 15, 17.5, 20, ...$ 4. Evaluate handwritten sum: $$\sum_{r=1}^{18} (3r + 2)$$ Step 1: Split the sum: $$\sum_{r=1}^{18} (3r + 2) = 3 \sum_{r=1}^{18} r + \sum_{r=1}^{18} 2$$ Step 2: Use formulas: $$\sum_{r=1}^n r = \frac{n(n+1)}{2}$$ So: $$3 \times \frac{18 \times 19}{2} + 2 \times 18 = 3 \times 171 + 36 = 513 + 36 = 549$$ Final answers: 1.a $d=4$ 1.b $a=6$ 1.c $S_{30} = 1920$ 2.i $x=2$ 2.ii Sequence: 4,5,6,... 2.iii 19th term = 22 3.a $d=2.5$ 3.b $a=12.5$ 3.c Sequence: $12.5, 15, 17.5, 20,...$ 4. Summation result: 549