Find the sum of the following geometric series:

Question:

Find the sum of the following geometric series:

(i) 0.15 + 0.015 + 0.0015 + ... to 8 terms;

(ii) $\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2 \sqrt{2}}+\ldots$ to 8 terms;

(iii) $\frac{2}{9}-\frac{1}{3}+\frac{1}{2}-\frac{3}{4}+\ldots$ to 5 terms;

(iv) (x +y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ... to n terms;

(v) $\frac{3}{5}+\frac{4}{5^{2}}+\frac{3}{5^{3}}+\frac{4}{5^{4}}+\ldots$ to $2 n$ terms;

(vi) $\frac{a}{1+i}+\frac{a}{(1+i)^{2}}+\frac{a}{(1+i)^{3}}+\ldots+\frac{a}{(1+i)^{n}}$.

(vii) 1, −aa2, −a3, ... to n terms (a ≠ 1)

(viii) x3x5x7, ... to n terms

(ix) $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$ to $n$ terms

 

 

Solution:

(i) Here, $a=0.15$ and $r=\frac{a_{2}}{a_{1}}=\frac{0.015}{0.15}=\frac{1}{10}$.

$\mathrm{S}_{8}=a\left(\frac{1-r^{8}}{1-r}\right)$

$=0.15\left(\frac{1-\left(\frac{1}{10}\right)^{8}}{1-\frac{1}{10}}\right)$

$=0.15\left(\frac{1-\frac{1}{10^{8}}}{\frac{1}{10}}\right)$

$=\frac{1}{6}\left(1-\frac{1}{10^{8}}\right)$

(ii) Here, $a=\sqrt{2}$ and $r=\frac{1}{2}$.

$\mathrm{S}_{8}=\mathrm{a}\left(\frac{1-\mathrm{r}^{8}}{1-\mathrm{r}}\right)$

$=\sqrt{2}\left(\frac{1-\left(\frac{1}{2}\right)^{8}}{1-\frac{1}{2}}\right)$

$=\sqrt{2}\left(\frac{1-\frac{1}{256}}{\frac{1}{2}}\right)$

$=2 \sqrt{2}\left(\frac{255}{256}\right)$

$=\frac{255 \sqrt{2}}{128}$

(iii) Here, $\mathrm{a}=\frac{2}{9}$ and $r=-\frac{3}{2}$.

$S_{5}=a\left(\frac{r^{5}-1}{r-1}\right)$

$=\frac{2}{9}\left(\frac{\left(\frac{-3}{2}\right)^{5}-1}{\frac{-3}{2}-1}\right)$

$=\frac{2}{9}\left(\frac{\left(-\frac{243}{32}\right)-1}{\frac{-3}{2}-1}\right)$

$=\frac{2}{9}\left(\frac{\frac{-275}{\frac{32}{-5}}}{2}\right)$

$=\frac{1100}{1440}$

$=\frac{55}{72}$

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