# How do you find the sum of the infinite geometric series 27+9+3+1+...?

$40.5$

#### Explanation:

The given infinite geometric series

$27 + 9 + 3 + 1 + \setminus \ldots$

has first term $a = 27$ & the common ratio $r = \frac{9}{27} = \frac{3}{9} = \setminus \ldots = \frac{1}{3}$ h

Hence the sum of given infinite G.P.

$= \frac{a}{1 - r}$

$= \setminus \frac{27}{1 - \frac{1}{3}}$

$= \frac{81}{2}$

$= 40.5$

Jul 11, 2018

${S}_{\infty} = \frac{81}{2}$

#### Explanation:

$\text{the sum to infinity of a geometric series is}$

•color(white)(x)S_oo =a/(1-r)to -1 < r <1

$\text{where a is the first term and r the common ratio}$

$\text{here "a=27" and } r = \frac{9}{27} = \frac{3}{9} = \frac{1}{3}$

${S}_{\infty} = \frac{27}{1 - \frac{1}{3}} = \frac{27}{\frac{2}{3}} = \frac{81}{2}$

Jul 11, 2018

The answer is $= 40.5$

#### Explanation:

If the common ratio of a geometric series is

$| r | < 1$

Then,

The sum of the infinite geometric series

${a}_{1} + {a}_{1} r + {a}_{1} {r}^{2} + \ldots \ldots . . a - 1 {r}^{n} + \ldots \ldots .$

is given by .

${S}_{\infty} = {a}_{1} / \left(1 - r\right)$

Here the series is

$27 + 27 \cdot \frac{1}{3} + 27 \cdot {\left(\frac{1}{3}\right)}^{2} + 27 \cdot \left(\frac{1}{3} ^ 3\right) + \ldots \ldots \ldots \ldots \ldots . .$

${a}_{1} = 27$

and

$r = \frac{1}{3}$

${S}_{\infty} = \frac{27}{1 - \frac{1}{3}} = \frac{27}{\frac{2}{3}} = 40.5$