Gp Exercise 7 4 1Da668

1. **Find the common ratio $r$ for each GP:** (i) Sequence: $2, 8, 32, 128, \ldots$ Common ratio formula: $r = \frac{a_2}{a_1}$ $r = \frac{8}{2} = 4$ (ii) Sequence: $99, 33, 11, \ldots$ $r = \frac{33}{99} = \frac{1}{3}$ (iii) Sequence: $\sqrt{2}, \sqrt{6}, 3\sqrt{2}, \ldots$ $r = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$ (iv) Sequence: $0.1, 0.02, 0.004, 0.0008, \ldots$ $r = \frac{0.02}{0.1} = 0.2$ (v) Sequence: $k, 2k^2, 4k^3, \ldots$ $r = \frac{2k^2}{k} = 2k$ (vi) Sequence: $-1, \frac{1}{3}, -\frac{1}{9}, \frac{1}{27}, \ldots$ $r = \frac{\frac{1}{3}}{-1} = -\frac{1}{3}$ 2. **Check if sequences are GPs and find $r$ and general term $a_n$:** (i) $4, 8, 16, \ldots$ $r = \frac{8}{4} = 2$, GP confirmed. General term: $a_n = 4 \times 2^{n-1}$ (ii) $1, 5, 8, 12, 15, \ldots$ Ratios: $\frac{5}{1} = 5$, $\frac{8}{5} = 1.6$, not constant, not GP. (iii) $9, 3, 1, \ldots$ $r = \frac{3}{9} = \frac{1}{3}$, GP confirmed. General term: $a_n = 9 \times \left(\frac{1}{3}\right)^{n-1}$ (iv) $11, 9, 7, 5, \ldots$ Ratios: $\frac{9}{11} \neq \frac{7}{9}$, not constant, not GP. (v) $12, 3, \frac{3}{4}, \ldots$ $r = \frac{3}{12} = \frac{1}{4}$, check next: $\frac{3/4}{3} = \frac{1}{4}$, GP confirmed. General term: $a_n = 12 \times \left(\frac{1}{4}\right)^{n-1}$ (vi) $10, 1, \frac{1}{10}, \frac{1}{100}, \ldots$ $r = \frac{1}{10} = 0.1$, check next: $\frac{1/10}{1} = 0.1$, GP confirmed. General term: $a_n = 10 \times (0.1)^{n-1}$ 3. **Find sum of GPs:** Sum formula: $$S_n = a \frac{1-r^n}{1-r}$$ for $r \neq 1$ (i) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ to 9 terms $a = \frac{1}{3}, r = \frac{1}{3}, n=9$ $$S_9 = \frac{1}{3} \times \frac{1-(\frac{1}{3})^9}{1-\frac{1}{3}} = \frac{1}{3} \times \frac{1-\frac{1}{19683}}{\frac{2}{3}} = \frac{1}{3} \times \frac{19682/19683}{2/3} = \frac{1}{3} \times \frac{19682}{19683} \times \frac{3}{2} = \frac{19682}{39366} \approx 0.5$$ (ii) $1, 5, 25, 125, \ldots$ to 7 terms $a=1, r=5, n=7$ $$S_7 = 1 \times \frac{1-5^7}{1-5} = \frac{1-78125}{-4} = \frac{-78124}{-4} = 19531$$ (iii) $\frac{1}{5}, \frac{1}{10}, \frac{1}{20}, \ldots$ to 10 terms $a=\frac{1}{5}, r=\frac{1}{2}, n=10$ $$S_{10} = \frac{1}{5} \times \frac{1-(\frac{1}{2})^{10}}{1-\frac{1}{2}} = \frac{1}{5} \times \frac{1-\frac{1}{1024}}{\frac{1}{2}} = \frac{1}{5} \times \frac{1023/1024}{1/2} = \frac{1}{5} \times \frac{1023}{1024} \times 2 = \frac{2046}{5120} \approx 0.4$$ (iv) $400, 200, 100, \ldots$ to 5 terms $a=400, r=\frac{200}{400} = \frac{1}{2}, n=5$ $$S_5 = 400 \times \frac{1-(\frac{1}{2})^5}{1-\frac{1}{2}} = 400 \times \frac{1-\frac{1}{32}}{\frac{1}{2}} = 400 \times \frac{31/32}{1/2} = 400 \times \frac{31}{32} \times 2 = 775$$ (v) $16, 8, 4, 2, \ldots$ to 8 terms $a=16, r=\frac{8}{16} = \frac{1}{2}, n=8$ $$S_8 = 16 \times \frac{1-(\frac{1}{2})^8}{1-\frac{1}{2}} = 16 \times \frac{1-\frac{1}{256}}{\frac{1}{2}} = 16 \times \frac{255/256}{1/2} = 16 \times \frac{255}{256} \times 2 = 31.875$$ (vi) $-2, 1, -\frac{1}{2}, \frac{1}{4}, \ldots$ to 12 terms $a=-2, r=\frac{1}{-2} = -\frac{1}{2}, n=12$ $$S_{12} = -2 \times \frac{1-(-\frac{1}{2})^{12}}{1-(-\frac{1}{2})} = -2 \times \frac{1-\frac{1}{4096}}{1+\frac{1}{2}} = -2 \times \frac{4095/4096}{\frac{3}{2}} = -2 \times \frac{4095}{4096} \times \frac{2}{3} = -\frac{8190}{12288} = -0.666\ldots$$ 4. **Sum to $n$ terms and find $S_8$ for $4, 2, 1, \frac{1}{2}, \ldots$** $a=4, r=\frac{2}{4} = \frac{1}{2}$ Sum formula: $$S_n = 4 \times \frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}} = 4 \times \frac{1-(\frac{1}{2})^n}{\frac{1}{2}} = 8 \times \left(1 - \left(\frac{1}{2}\right)^n\right)$$ Calculate $S_8$: $$S_8 = 8 \times \left(1 - \left(\frac{1}{2}\right)^8\right) = 8 \times \left(1 - \frac{1}{256}\right) = 8 \times \frac{255}{256} = \frac{2040}{256} = 7.96875$$ Check with calculator: $7.96875$ matches the sum.