1. The problem states that the sum of the first 10 terms of an arithmetic progression (A.P.) is 240 and the 8th term is 34. Step 1: Recall the sum formula for the first n terms of an A.P.: $$S_n = \frac{n}{2} [2a + (n-1)d]$$ Step 2: Let first term be $a$ and common difference be $d$. Given $S_{10} = 240$, so: $$240 = \frac{10}{2} [2a + 9d] = 5(2a + 9d)$$ Simplify: $$2a + 9d = \frac{240}{5} = 48 \quad ...(1)$$ Step 3: The 8th term $T_8$ is given by: $$T_8 = a + 7d = 34 \quad ...(2)$$ Step 4: From (2), express $a$: $$a = 34 - 7d$$ Step 5: Substitute $a$ in (1): $$2(34 - 7d) + 9d = 48$$ $$68 - 14d + 9d = 48$$ $$68 - 5d = 48$$ $$-5d = 48 - 68 = -20$$ $$d = \frac{20}{5} = 4$$ Step 6: Find $a$: $$a = 34 - 7(4) = 34 - 28 = 6$$ Step 7: Find the sum of the first 30 terms $S_{30}$: $$S_{30} = \frac{30}{2} [2a + 29d] = 15 [2(6) + 29(4)] = 15 [12 + 116] = 15 \times 128 = 1920$$ 2. Given the first three terms of an A.P. are $2x$, $(3x - 1)$, and $x + 4$. Step 1: For an A.P., the difference between consecutive terms is constant: $$(3x - 1) - 2x = (x + 4) - (3x -1)$$ Simplify the left: $$3x - 1 - 2x = x - 1$$ Simplify the right: $$x + 4 - 3x + 1 = -2x + 5$$ Step 2: Set equal: $$x - 1 = -2x + 5$$ $$x + 2x = 5 + 1$$ $$3x = 6$$ $$x = 2$$ Step 3: Find the sequence terms by substituting $x = 2$: $$1^{st} = 2(2) = 4$$ $$2^{nd} = 3(2) - 1 = 6 - 1 = 5$$ $$3^{rd} = 2 + 4 = 6$$ So the sequence starts as 4, 5, 6, ... Step 4: Find the common difference: $$d = 5 - 4 = 1$$ Step 5: Find the 19th term $T_{19}$: $$T_{19} = a + 18d = 4 + 18(1) = 22$$ 3. The 4th term of an A.P. is 20, and the 8th term is 1.5 times the 4th term. Step 1: Write terms explicitly: $$T_4 = a + 3d = 20$$ $$T_8 = a + 7d = 1.5 \times 20 = 30$$ Step 2: Subtract the two equations: $$(a + 7d) - (a + 3d) = 30 - 20$$ $$4d = 10$$ $$d = \frac{10}{4} = 2.5$$ Step 3: Find $a$ using $T_4$: $$a + 3(2.5) = 20$$ $$a + 7.5 = 20$$ $$a = 20 - 7.5 = 12.5$$ Step 4: Write the sequence: $$T_n = a + (n-1)d = 12.5 + (n-1)(2.5)$$ 4. Evaluate the summation: $$\sum_{r=1}^{18} (3r + 2)$$ Step 1: Split the summation: $$\sum_{r=1}^{18} 3r + \sum_{r=1}^{18} 2 = 3 \sum_{r=1}^{18} r + 2 \times 18$$ Step 2: Sum of first 18 natural numbers: $$\sum_{r=1}^{18} r = \frac{18 \times 19}{2} = 171$$ Step 3: Calculate: $$3 \times 171 + 36 = 513 + 36 = 549$$