Question

Evaluate the series $12 + 6 + 3...$ .

Answer 1

To evaluate the series $12 + 6 + 3...$ . It is an infinite series so we have to evaluate the general summation for this series.
First classify the type of series then apply the formula according to that.
Some formulae to remember:
When $|r| < 1$ , then the sum of the infinite series converges to $\sum\limits_{n = 1}^\infty {a{r^{n - 1}}} = \dfrac{a}{{1 - r}}$ ,where, ‘a’ is the first term and ‘r’ is the common ratio.
When $r = 1$ , which means each term of the series is the same and the series is infinite, hence, the series does not converge.
When $|r| > 1$ then the terms of the series become larger and larger and hence the sum becomes larger and larger, thus the sum of the infinite series does not exist.

Complete step-by-step answer:
Given series $12 + 6 + 3...$
To evaluate the given series, first we need to know the type of series.
We have, $6 - 12 = - 6$ , $3 - 6 = - 3$ and so on
Since, the common difference between the terms is not same, so, it is not an arithmetic progression.
Now, $\dfrac{6}{{12}} = \dfrac{1}{2}$ , $\dfrac{3}{6} = \dfrac{1}{2}$ and so on.
Since, the common ratio of the terms are same, therefore, it is a geometric series.
Thus, the common ratio of the above series $r = \dfrac{1}{2}$ and the first term is $a = 12$ .
Now, we know the general form of the geometric series is $\sum\limits_{n = 1}^\infty {a{r^{n - 1}}}$ , where, ‘n’ represents the ${n^{th}}$ term.
Now, since, $|r| < 1$ , therefore, sum of the infinite series is given by $\sum\limits_{n = 1}^\infty {a{r^{n - 1}}} = \dfrac{a}{{1 - r}}$ .
Putting the values of ‘a’ and ‘r’, we get,
$\sum\limits_{n = 1}^\infty {12{{\left( {\dfrac{1}{2}} \right)}^{n - 1}}} = \dfrac{{12}}{{1 - \dfrac{1}{2}}}$
On solving, we get,
$\dfrac{{12}}{{1 - \dfrac{1}{2}}} = \dfrac{{12}}{{\dfrac{{2 - 1}}{2}}}$
$\dfrac{{12}}{{\dfrac{{2 - 1}}{2}}} = \dfrac{{12 \times 2}}{1}$
i.e., $\sum\limits_{n = 1}^\infty {12{{\left( {\dfrac{1}{2}} \right)}^{n - 1}}} = 24$ .

Note: First step is to check the type of series, whether it is finite or infinite or it is in AP or in GP. According to that formula for summation will follow.
Check the value of $r$ , whether it is greater than, smaller than or equal to $1$ and according to that solve further.
If r is greater than 1, then, it means the terms are becoming larger and larger and hence the sum becomes larger and hence that is why sum of such infinite series does not exist.