Find the sum of the following geometric progressions:

Question:

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.

Solution:

(i) Here, = 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 $r=\frac{1}{2}$

$\therefore S_{n}=a\left(\frac{1-r^{10}}{1-r}\right)$

$=4\left(\frac{1-\left(\frac{1}{2}\right)^{10}}{1-\left(\frac{1}{2}\right)}\right)$

$=4\left(\frac{1-\left(\frac{1}{1024}\right)}{\frac{1}{2}}\right)$

$=8\left(1-\frac{1}{1024}\right)$

$=\frac{1023}{128}$

Leave a comment

Click here to get exam-ready with eSaral