Identify whether the following sequence is a geometric sequence or not.
\[\dfrac{1}{2},\dfrac{2}{4},\dfrac{4}{8},\dfrac{8}{{16}}\]

Hint- Each term of a geometric sequence increases or decreases by a constant factor called the common ratio.
Let ${a_1},{a_2},...,{a_n}$ be a geometric sequence.
Common ratio is of this geometric sequence found using the following formula:
$r = \dfrac{{{a_n}}}{{{a_{n - 1}}}}{\text{ where }}n > 1$

Complete step by step answer:
To find whether the given sequence is a geometric sequence are not we should check whether the sequence increases or decreases under a common ratio.
Let us divide each term by the previous term to determine whether a common ratio exists or not.
Here \[{a_1} = \dfrac{1}{2}\] and \[{a_2} = \dfrac{2}{4}\]
We can find common ratio by the formula,
\[\dfrac{{{a_2}}}{{{a_1}}} = r\]
We get,
\[\dfrac{{\dfrac{2}{4}}}{{\dfrac{1}{2}}} = 1\]
Common ratio \[r = 1\]
Now let us consider the next two terms in the given sequence
Again, \[{a_2} = \dfrac{2}{4}\]and \[{a_3} = \dfrac{4}{8}\]
We can find common ratio by the formula
\[\dfrac{{{a_3}}}{{{a_2}}} = r\]
We get,
 \[\dfrac{{\dfrac{4}{8}}}{{\dfrac{2}{4}}} = 1\]
Common ratio \[r = 1\]
Now let us consider the final two terms in the sequence to find the common ratio.
 \[{a_3} = \dfrac{4}{8}\] and \[{a_4} = \dfrac{8}{{16}}\]
We can find common ratio by the formula,
\[\dfrac{{{a_4}}}{{{a_3}}} = r\]
We get,
\[\dfrac{{\dfrac{8}{{16}}}}{{\dfrac{4}{8}}} = 1\]
Common ratio \[r = 1\]
In the given sequence the common ratio between every term is found to be one, therefore from the definition of a geometric sequence we can assure that the given sequence is a geometric sequence.
Hence
The given sequence is a geometric sequence with the common ratio is 1.

Note: The number multiplied or divided at each stage of a geometric sequence is called the "common ratio" r, because if we divide that is, if we find the ratio of successive terms, we will always get this common value.