Two geometric means between the number $1$ and $64$ are
A. $1$ and $64$
B. $4$ and $16$
C. $2$ and $16$
D. $2$ and $8$

Hint: To understand geometric means we need to first understand geometric progression. For three numbers to be in geometric progression, they have to have a common ratio that each term is multiplied with to get the next term. To get the answer for this question, we need to insert two geometric means between the given numbers.

Formula used: If these three terms are in a geometric progress then the middle term is the geometric mean of the first and third term. It is represented as
$GM(b) = \sqrt {ac} $
Where $a,b,c$ are in a GP
If $'n'$’ number of Geometric means are inserted between two numbers $a$ and $b$
Then $a,{G_1},{G_2},{G_3}...{G_n},b$ form a geometric progression where
First term $ = a$
${\left( {n + 2} \right)^{th}}$ term $ = b$
that is, $b = a{r^{(n + 2 - 1)}} = a{r^{(n + 1)}}$
$ \Rightarrow \dfrac{b}{a} = {r^{(n + 1)}}$
$ \Rightarrow \dfrac{b}{a} = {r^{(n + 1)}}$
$ \Rightarrow r = {(\dfrac{b}{a})^{\dfrac{1}{{n + 1}}}}$
Therefore,
${G_n} = a{(\dfrac{b}{a})^{\dfrac{n}{{n + 1}}}}$

Complete step-by-step answer:
In the given question, it asks for two geometric means between the given numbers. Therefore
$n = 2$
This implies that we need to find ${G_1}$ and ${G_2}$ for a geometric progression where the first term $a = 1$ and the last term $b = 64$
From the equation for common ratio of these geometric progressions, the common ratio is
$ \Rightarrow r = {(\dfrac{{64}}{1})^{\dfrac{1}{{2 + 1}}}}$
$ \Rightarrow r = \sqrt[3]{{64}} = 4$
Using the formula for the geometric means inserted between two numbers, we get that
${G_1} = 1{(4)^1} = 4$
Similarly, for the second geometric mean inserted between these numbers,
${G_2} = 1{(4)^2} = 16$
Therefore the two geometric means between the numbers $1$ and $64$ are $4$ and $16$

Thus the correct option is $(b)4$ and $16$.

Note:
This question can also be solved using the formula of geometric mean for three numbers. We first use the given numbers to find the middle term. Then we use the middle term as the first term and the last term as the last term to find the second middle term.