Let $$a_1, a_2, \ldots, a_{10}$$ be a G.P. If $$\frac{a_3}{a_1} = 25$$, then $$\frac{a_9}{a_5}$$ equals:

We have a geometric progression whose first ten terms are $$a_1, a_2, \ldots , a_{10}$$. In any G.P., every term can be written using the first term and the common ratio. If we denote the common ratio by $$r$$, then the general formula is

$$a_n = a_1 \, r^{\,n-1}\;.$$

Using this formula for the third term, we get

$$a_3 = a_1 \, r^{\,3-1}= a_1 \, r^2.$$

Now we look at the given condition $$\dfrac{a_3}{a_1}=25$$. Substituting the expression for $$a_3$$, we obtain

$$\frac{a_3}{a_1}= \frac{a_1 \, r^2}{a_1}= r^2 = 25.$$

So we have the equation $$r^2 = 25$$. Taking the square root on both sides,

$$r = \pm 5.$$

Next, we want $$\dfrac{a_9}{a_5}$$. Using the general formula again,

$$a_9 = a_1 \, r^{\,9-1}= a_1 \, r^8,$$

$$a_5 = a_1 \, r^{\,5-1}= a_1 \, r^4.$$

Hence,

$$\frac{a_9}{a_5}= \frac{a_1 \, r^8}{a_1 \, r^4}= r^{\,8-4}= r^4.$$

We already know $$r^2 = 25$$, so

$$r^4 = (r^2)^2 = 25^2 = 625.$$

Since $$625 = 5^4$$ (and this value is the same whether $$r=5$$ or $$r=-5$$ because the fourth power removes the sign), we conclude

$$\frac{a_9}{a_5}=5^4.$$

Hence, the correct answer is Option A.