Let ${{a}_{1}},{{a}_{2}},\ldots \ldots ,{{a}_{10}}$ be a G.P. If $\dfrac{{{a}_{3}}}{{{a}_{1}}}=25$, then $\dfrac{{{a}_{9}}}{{{a}_{5}}}$ equals :
A. $2\left( {{5}^{2}} \right)$
B. $4\left( {{5}^{2}} \right)$
C. ${{5}^{4}}$
D. ${{5}^{3}}$

Hint: We have been given here that ${{a}_{1}},{{a}_{2}},\ldots \ldots ,{{a}_{10}}$ are in G.P along with a relation, $\dfrac{{{a}_{3}}}{{{a}_{1}}}=25$. We will use the concept of general term of G.P, given by $a{{r}^{n-1}}$, taking ${{a}_{1}}$ as $a$ and ${{a}_{2}}=ar$. Then we will substitute the value of the term in the given relation and get the answer.

Complete step-by-step answer:
In the question, it is mentioned that, ${{a}_{1}},{{a}_{2}},\ldots \ldots ,{{a}_{10}}$ are in G.P and that $\dfrac{{{a}_{3}}}{{{a}_{1}}}=25$. We have been asked to find the value of $\dfrac{{{a}_{9}}}{{{a}_{5}}}$. We have to understand the concept of G.P here. G.P is a sequence of numbers where each term after the first term is found by multiplying the previous one by a fixed, non-zero number, which is called the common ratio. The general term of a G.P is denoted as, $a{{r}^{n-1}}$, where $a$ is the first term and $r$ is the common ratio. So, according to the general formula, we can write,
$\begin{align}
  & {{a}_{3}}=a{{r}^{3-1}}\Rightarrow {{a}_{3}}=a{{r}^{2}} \\
 & {{a}_{1}}=a{{r}^{1-1}}\Rightarrow {{a}_{1}}=a \\
\end{align}$
Now, we have been given $\dfrac{{{a}_{3}}}{{{a}_{1}}}=25$, therefore substituting the terms in it, we get,
$\begin{align}
  & \dfrac{a{{r}^{2}}}{a}=25 \\
 & \Rightarrow {{r}^{2}}=25 \\
 & \Rightarrow r=\pm 5 \\
\end{align}$
We will consider the common ratio as + 5. So, now we have to find the ratio of ${{a}_{9}}$ and ${{a}_{5}}$. So, we will start by writing the general terms for ${{a}_{9}}$ and ${{a}_{5}}$. So, we will get them as,
$\begin{align}
  & {{a}_{9}}=a{{r}^{9-1}}\Rightarrow {{a}_{9}}=a{{r}^{8}} \\
 & {{a}_{5}}=a{{r}^{5-1}}\Rightarrow {{a}_{5}}=a{{r}^{4}} \\
\end{align}$
So, we will substitute the terms in the given relation, $\dfrac{{{a}_{9}}}{{{a}_{5}}}$, and get,
$\begin{align}
  & \dfrac{{{a}_{9}}}{{{a}_{5}}}=\dfrac{a{{r}^{8}}}{a{{r}^{4}}} \\
 & \Rightarrow \dfrac{{{a}_{9}}}{{{a}_{5}}}={{r}^{8-4}} \\
 & \Rightarrow \dfrac{{{a}_{9}}}{{{a}_{5}}}={{r}^{4}} \\
\end{align}$
We have obtained the value of $r=5$, so we will substitute this value in the above equality and get,
$\dfrac{{{a}_{9}}}{{{a}_{5}}}={{5}^{4}}$

So, the correct answer is “Option c”.

Note: The general term of G.P helps in finding the common ratio from the given data. There is no need of finding the first term of the G.P as we can see that the first term considered as a gets cancelled out and we need the common ratio at the end. This concept can be used in any ratio solving problems. Even if we use the value of $r=-5$, we will get the same answer as the power is an even number, that is, ${{\left( -5 \right)}^{4}}={{\left( 5 \right)}^{4}}={{5}^{4}}$.