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Question

# Find the sum of the following geometric series: (i) 0.15 + 0.015 + 0.0015 + ... to 8 terms; (ii) $\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+...\mathrm{to}8\mathrm{terms};$ (iii) $\frac{2}{9}-\frac{1}{3}+\frac{1}{2}-\frac{3}{4}+...\mathrm{to}5\mathrm{terms};$ (iv) (x +y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ... to n terms; (v) $\frac{3}{5}+\frac{4}{{5}^{2}}+\frac{3}{{5}^{3}}+\frac{4}{{5}^{4}}+...\mathrm{to}2n\mathrm{terms};$ (vi) $\frac{a}{1+i}+\frac{a}{\left(1+i{\right)}^{2}}+\frac{a}{\left(1+i{\right)}^{3}}+...+\frac{a}{\left(1+i{\right)}^{n}}.$ (vii) 1, −a, a2, −a3, ... to n terms (a ≠ 1) (viii) x3, x5, x7, ... to n terms (ix) $\sqrt{7},\sqrt{21},3\sqrt{7},...\mathrm{to}n\mathrm{terms}$

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Solution

## (i) Here, a = 0.15 and r $=\frac{{a}_{2}}{{a}_{1}}=\frac{0.015}{0.15}=\frac{1}{10}$. ${\mathrm{S}}_{8}=a\left(\frac{1-{r}^{8}}{1-r}\right)\phantom{\rule{0ex}{0ex}}=0.15\left(\frac{1-{\left(\frac{1}{10}\right)}^{8}}{1-\frac{1}{10}}\right)\phantom{\rule{0ex}{0ex}}=0.15\left(\frac{1-\frac{1}{{10}^{8}}}{\frac{1}{10}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\left(1-\frac{1}{{10}^{8}}\right)$ (ii) Here, a = $\sqrt{2}$ and r = $\frac{1}{2}$. ${\mathrm{S}}_{8}=\mathrm{a}\left(\frac{1-{\mathrm{r}}^{8}}{1-\mathrm{r}}\right)\phantom{\rule{0ex}{0ex}}=\sqrt{2}\left(\frac{1-{\left(\frac{1}{2}\right)}^{8}}{1-\frac{1}{2}}\right)\phantom{\rule{0ex}{0ex}}=\sqrt{2}\left(\frac{1-\frac{1}{256}}{\frac{1}{2}}\right)\phantom{\rule{0ex}{0ex}}=2\sqrt{2}\left(\frac{255}{256}\right)\phantom{\rule{0ex}{0ex}}=\frac{255\sqrt{2}}{128}$ (iii) Here, a =$\frac{2}{9}\mathrm{and}r=-\frac{3}{2}$. ${S}_{5}=a\left(\frac{{r}^{5}-1}{r-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{9}\left(\frac{{\left(\frac{-3}{2}\right)}^{5}-1}{\frac{-3}{2}-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{9}\left(\frac{\left(-\frac{243}{32}\right)-1}{\frac{-3}{2}-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{9}\left(\frac{\frac{-275}{32}}{\frac{-5}{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1100}{1440}\phantom{\rule{0ex}{0ex}}=\frac{55}{72}$

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