Question

How do you find the sum of the infinite geometric series given $18 - 12 + 8 - .....$?

Answer 1

Hint: In order to find the sum of the given infinite geometric series, we need to compare the given value with the general geometric series, finding the value of common ratio, then comparing it with $1$ and substitute the value in the given formula of sum for infinite series that is ${S_\infty } =\dfrac{a}{{1 - r}}$, where $\left| r \right| < 1$.

Complete step by step solution:
The series we are given is $18 - 12 + 8 - .....$.
Comparing with the general geometric series that is $a, ar, a{r^2}......,a{r^{n - 1}},.....$, we get first term $a = 18$, second term as $ar = - 12$ and so on.
Dividing first term by second term, we get:
$
  \dfrac{{second}}{{first}} = \dfrac{{ar}}{a} \\
  \dfrac{{ar}}{a} = \dfrac{{ - 12}}{{18}} \\
  r = \dfrac{{ - 12}}{{18}} = \dfrac{{ - 2}}{3} \\
 $
Since, $\left| r \right| = \left| {\dfrac{{ - 2}}{3}} \right| = \left| {\dfrac{2}{3}} \right| < 1$, so we use the formula ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio and ${S_\infty }$ is the sum of the infinite series.
Substituting the values obtained in the formula ${S_\infty } = \dfrac{a}{{1 - r}}$, and we get:
$
  {S_\infty } = \dfrac{a}{{1 - r}} = \dfrac{{18}}{{1 - \left( { - \dfrac{2}{3}} \right)}} \\
  {S_\infty } = \dfrac{{18}}{{1 + \dfrac{2}{3}}} = \dfrac{{18}}{{\dfrac{{3 + 2}}{3}}} = \dfrac{{18}}{{\dfrac{5}{3}}} = \dfrac{{18 \times 3}}{5} = \dfrac{{54}}{5} \\
 $
And, here is the sum of the geometric series we obtained.

Therefore, the sum of the infinite geometric series given $18 - 12 + 8 - .....$ is $\dfrac{{54}}{5}$.

Additional Information:
A finite series is a series whose last term is known, in other words the number of terms are known to us and is represented by $a, ar, a{r^2}......,a{r^n}$, where $a$ is the first term and $r$ is the common ratio.
An infinite geometric series sum is the sum of an infinite geometric sequence. And thus, we know that infinite means not having any last term, so infinite series does not have any last term, and the series is represented by $a, ar, a{r^2}......,a{r^{n - 1}},.....$, where $a$ is the first term and $r$ is the common ratio.

Note:
1) The values present should be considered as $18, - 12,8 - .....$ not as $18 - 12$ or $12 + 8$
2) There is a huge difference between infinite and finite geometric series, do not consider them the same.