Question

How do you find the sum of the infinite geometric series $12 + 6 + 3 + ....?$

Answer 1

Hint: The given series is an infinite geometric series and we know that the geometric series has a fixed ratio of their consecutive terms. To find the sum of infinite geometric series, first of all find the common ratio between terms, by dividing a term with its progressive term.

Formula used:
1) Common ratio of a G.P.: $r = \dfrac{{{u_{n + 1}}}}{{{u_n}}}$
2) Infinite sum of G.P.: ${S_\infty } = \dfrac{a}{{1 - r}}$, where ${S_\infty }$, $a$ and $r$ are the sum of infinite geometric series, first term and the common ratio of the series respectively.

Complete step by step solution:
In order to find the sum of the infinite geometric series $12 + 6 + 3 + ....$ we will first find the common ratio of the series as following
$r = \dfrac{{{u_{n + 1}}}}{{{u_n}}},\;{\text{where}}\;{u_{n + 1}}\;{\text{and}}\;{u_n}$ are “n+1th” and “nth” term of the geometric series respectively
We will take second and first term, to find the common ratio,
$ \Rightarrow r = \dfrac{6}{{12}} = \dfrac{1}{2}$
Now, we will use the formula for sum of infinite terms of geometric series which is given as follows
${S_\infty } = \dfrac{a}{{1 - r}},\;{\text{where}}\;{S_\infty },\;a\;{\text{and}}\;r$ are sum of infinite geometric series, first term of the series and common ratio of the series respectively
In the series $12 + 6 + 3 + ....$ the first term is $a = 12$
Putting $a = 12\;{\text{and}}\,r = \dfrac{1}{2}$ in the above formula, we will get
${S_\infty } = \dfrac{{12}}{{1 - \dfrac{1}{2}}} = \dfrac{{12}}{{2 - 1}} = 24$

Therefore the required infinite sum of the given series is equal to $24$.

Note:
The common ratio of geometric series or progression also informs us about the essence of the series, which either increases or decreases depending on whether the value of the common ratio is greater than one or less than one.