Question

How do you find the $\;8th$ term in this geometric sequence $\;8,4,2,1....$ ?

Answer 1

To solve such questions, first study the given sequence carefully and then try to find the common ratio. Then apply the formula to find the ${n^{th}}$ term of the series, by substituting the values in the formula after comparing them to the terms given in the question.

Formula used: The formula used to calculate and find the required term of a given geometric sequence is as given below:
${n^{th}}term = a{r^{n - 1}}$
Where $a$ is the initial term of the geometric series ( $a{r^n}$), $r$ is the common ratio of the series, and $n$ is the position of the term that has to be found out.

Complete step-by-step solution:
Given the geometric series as $\;8,4,2,1....$ , with the initial term $a$ as $a = 8$ .
To find the common ratio, divide any term, except for the first term, by the term before it, such as $\dfrac{{a{r^k}}}{{a{r^{k - 1}}}} = r$ .
In this sequence, $\;8,4,2,1....$ , divide the second term by the first to get the common ratio $r$ as,
$r = \dfrac{4}{8} = \dfrac{1}{2}$
Now, substitute the value of $a$ and $r$ in the formula $a{r^{n - 1}}$ to calculate the ${n^{th}}$ term of the series,
Here, the value of $n$ is $n = 8$ as we have to find the $\;8th$ term of the geometric sequence.
Substituting the values to get,
$\Rightarrow 8 \times {\left( {\dfrac{1}{2}} \right)^{8 - 1}}$
By simplifying the expression we get,
$\Rightarrow 8 \times {\left( {\dfrac{1}{2}} \right)^7}$
$\Rightarrow 8 \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$
Further simplify the expression, to get,
$= \dfrac{1}{{16}}$

Hence, the $\;8th$ term in the given geometric sequence $\;8,4,2,1....$ is $\dfrac{1}{{16}}$.

Note: Always remember to first find the common ratio $r$ and the first term $a$ of the geometric sequence given in the question. Also, keep in mind to write down the number of the term that has to be calculated from the question and then begin solving it by applying the correct formula.