Find the sum of the following geometric progressions:
(i) 2, 6, 18, ... to 7 terms;
(ii) 1, 3, 9, 27, ... to 8 terms;
(iii) 1, −1/2, 1/4, −1/8, ... to 9 terms;
(iv) $\left(a^{2}-b^{2}\right),(a-b),\left(\frac{a-b}{a+b}\right), \ldots$ to $n$ terms;
(v) 4, 2, 1, 1/2 ... to 10 terms.
(i) Here, a = 2 and r = 3.
$\therefore S_{7}=a\left(\frac{r^{7}-1}{r-1}\right)$
$=2\left(\frac{3^{7}-1}{3-1}\right)$
$=2187-1$
$=2186$
(ii) Here, a = 1 and r = 3.
$\therefore S_{8}=a\left(\frac{r^{8}-1}{r-1}\right)$
$=1\left(\frac{3^{8}-1}{3-1}\right)$
$=\frac{6561-1}{2}$
$=3280$
(iii) Here, $a=1$ and $r=-\frac{1}{2}$.
$\therefore S_{9}=a\left(\frac{1-r^{9}}{1-r}\right)$
$=1\left(\frac{1-\left(-\frac{1}{2}\right)^{9}}{1-\left(-\frac{1}{2}\right)}\right)$
$=\frac{1-\left(-\frac{1}{512}\right)}{\frac{3}{2}}$
$=\frac{\frac{513}{512}}{\frac{3}{2}}$
$=\frac{513 \times 2}{512 \times 3}$
$=\frac{171}{256}$
(iv) Here, $a=a^{2}-b^{2}$ and $r=\frac{1}{a+b}$.
$\therefore S_{n}=a\left(\frac{1-r^{n}}{1-r}\right)$
$=\left(a^{2}-b^{2}\right)\left(\frac{1-\left(\frac{1}{a+b}\right)^{n}}{1-\left(\frac{1}{a+b}\right)}\right)$
$=\left(a^{2}-b^{2}\right)\left(\frac{\left(\frac{(a+b)^{n}-1}{(a+b)^{n}}\right)}{\frac{(a+b)-1}{a+b}}\right)$
$\Rightarrow S_{n}=\frac{(a+b)(a-b)}{(a+b)^{n-1}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right)$
$=\frac{(a-b)}{(a+b)^{n-2}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right)$
(v) Here, $a=4$ and $r=\frac{1}{2}$
$\therefore S_{n}=a\left(\frac{1-r^{10}}{1-r}\right)$
$=4\left(\frac{1-\left(\frac{1}{2}\right)^{10}}{1-\left(\frac{1}{2}\right)}\right)$
$=4\left(\frac{1-\left(\frac{1}{1024}\right)}{\frac{1}{2}}\right)$
$=8\left(1-\frac{1}{1024}\right)$
$=\frac{1023}{128}$
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