1. **Find the common difference $d$ for each arithmetic progression (AP).** The common difference formula is: $$d = a_2 - a_1$$ (i) For $2, 7, 12, \ldots$: $$d = 7 - 2 = 5$$ (ii) For $3, 1, -1, \ldots$: $$d = 1 - 3 = -2$$ (iii) For $1, 0.75, 0.5, \ldots$: $$d = 0.75 - 1 = -0.25$$ (iv) For $-2, -9, -16, \ldots$: $$d = -9 - (-2) = -9 + 2 = -7$$ (v) For $y, y + 3x, y + 6x, \ldots$: $$d = (y + 3x) - y = 3x$$ (vi) For $2x - 3y, 5x - y, 8x + y, \ldots$: $$d = (5x - y) - (2x - 3y) = 5x - y - 2x + 3y = 3x + 2y$$ 2. **Find the $n^{th}$ term $a_n$ of the APs.** The formula for the $n^{th}$ term is: $$a_n = a_1 + (n - 1)d$$ (i) $3, 8, 13, \ldots$ with $a_1=3$, $d=8-3=5$: $$a_n = 3 + (n-1)5 = 3 + 5n - 5 = 5n - 2$$ (ii) $1, -2, -5, \ldots$ with $a_1=1$, $d=-2 - 1 = -3$: $$a_n = 1 + (n-1)(-3) = 1 - 3n + 3 = 4 - 3n$$ (iii) $\frac{1}{2}, \frac{7}{6}, \frac{11}{6}, \ldots$ with $a_1=\frac{1}{2}$, $d=\frac{7}{6} - \frac{1}{2} = \frac{7}{6} - \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$: $$a_n = \frac{1}{2} + (n-1)\frac{2}{3} = \frac{1}{2} + \frac{2}{3}n - \frac{2}{3} = \frac{2}{3}n - \frac{1}{6}$$ (iv) $0.2, 1.1, 2, \ldots$ with $a_1=0.2$, $d=1.1 - 0.2 = 0.9$: $$a_n = 0.2 + (n-1)0.9 = 0.2 + 0.9n - 0.9 = 0.9n - 0.7$$ (v) $1, 3, 5, 7, \ldots$ with $a_1=1$, $d=3 - 1 = 2$: $$a_n = 1 + (n-1)2 = 1 + 2n - 2 = 2n - 1$$ (vi) $2, 4, 6, 8, \ldots$ with $a_1=2$, $d=4 - 2 = 2$: $$a_n = 2 + (n-1)2 = 2 + 2n - 2 = 2n$$ 3. **Find the indicated terms in the APs using $a_n = a_1 + (n-1)d$.** (i) $7, 11, 15, \ldots$ with $a_1=7$, $d=11 - 7 = 4$: $$a_6 = 7 + (6-1)4 = 7 + 20 = 27$$ $$a_{15} = 7 + (15-1)4 = 7 + 56 = 63$$ (ii) $\frac{1}{2}, \frac{5}{6}, \frac{7}{6}, \ldots$ with $a_1=\frac{1}{2}$, $d=\frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$: $$a_7 = \frac{1}{2} + (7-1)\frac{1}{3} = \frac{1}{2} + 2 = \frac{5}{2}$$ $$a_{12} = \frac{1}{2} + (12-1)\frac{1}{3} = \frac{1}{2} + \frac{11}{3} = \frac{3}{6} + \frac{22}{6} = \frac{25}{6}$$ (iii) $2.3, 3.7, 5.1, \ldots$ with $a_1=2.3$, $d=3.7 - 2.3 = 1.4$: $$a_9 = 2.3 + (9-1)1.4 = 2.3 + 11.2 = 13.5$$ $$a_{13} = 2.3 + (13-1)1.4 = 2.3 + 16.8 = 19.1$$ (iv) $9, 6, 3, \ldots$ with $a_1=9$, $d=6 - 9 = -3$: $$a_7 = 9 + (7-1)(-3) = 9 - 18 = -9$$ $$a_{20} = 9 + (20-1)(-3) = 9 - 57 = -48$$ (v) $y, y + 4x, 5x + y, y + 6x, \ldots$ with $a_1 = y$, $d = (y + 4x) - y = 4x$: $$a_5 = y + (5-1)4x = y + 16x$$ $$a_{10} = y + (10-1)4x = y + 36x$$ (vi) $\frac{p+q}{3}, \frac{p-q}{3}, \frac{p-3q}{3}, \ldots$ with $a_1=\frac{p+q}{3}$, $d=\frac{p-q}{3} - \frac{p+q}{3} = \frac{p-q - p - q}{3} = \frac{-2q}{3}$: $$a_5 = \frac{p+q}{3} + (5-1)\left(-\frac{2q}{3}\right) = \frac{p+q}{3} - \frac{8q}{3} = \frac{p+q - 8q}{3} = \frac{p - 7q}{3}$$ $$a_9 = \frac{p+q}{3} + (9-1)\left(-\frac{2q}{3}\right) = \frac{p+q}{3} - \frac{16q}{3} = \frac{p + q - 16q}{3} = \frac{p - 15q}{3}$$