If the sum of an infinite GP, $$a, ar, ar^2, ar^3, \ldots$$ is 15 and the sum of the squares of its each term is 150, then the sum of $$ar^2, ar^4, ar^6, \ldots$$ is:

Let us denote the first term of the given GP by $$a$$ and its common ratio by $$r$$. Because the series is infinite and is specified to be a GP, we necessarily have $$|r| < 1$$ so that all its sums converge.

We are told that the sum of the series

$$a + ar + ar^2 + ar^3 + \ldots$$

is $$15$$. For an infinite GP, the standard formula for the sum is

$$S_\infty \;=\; \frac{\text{first term}}{1 - \text{common ratio}}.$$

Using this formula we obtain

$$\frac{a}{1 - r} \;=\; 15.$$

Next, consider the series formed by squaring every term of the original GP:

$$a^2 + (ar)^2 + (ar^2)^2 + (ar^3)^2 + \ldots \;=\; a^2 + a^2 r^2 + a^2 r^4 + a^2 r^6 + \ldots$$

This, too, is an infinite GP with first term $$a^2$$ and common ratio $$r^2$$. Again invoking the infinite-sum formula, we have

$$\frac{a^2}{1 - r^2} \;=\; 150.$$

We thus possess the two equations

$$\frac{a}{1 - r} \;=\; 15 \quad\text{and}\quad \frac{a^2}{1 - r^2} \;=\; 150.$$

First, isolate $$a$$ from the first equation:

$$a \;=\; 15(1 - r).$$

Now square this expression so that we can substitute it into the second equation:

$$a^2 \;=\; \bigl[15(1 - r)\bigr]^2 \;=\; 225(1 - r)^2.$$

The second equation tells us simultaneously that

$$a^2 \;=\; 150(1 - r^2) \;=\; 150(1 - r)(1 + r).$$

Set the two expressions for $$a^2$$ equal:

$$225(1 - r)^2 \;=\; 150(1 - r)(1 + r).$$

Because $$|r| < 1$$ we have $$1 - r \neq 0$$, so we may safely divide both sides by $$1 - r$$ to obtain

$$225(1 - r) \;=\; 150(1 + r).$$

Expand and collect like terms:

$$225 - 225r \;=\; 150 + 150r.$$

Bring all terms to one side:

$$225 - 150 \;=\; 225r + 150r,$$

$$75 \;=\; 375r.$$

Solve for $$r$$:

$$r \;=\; \frac{75}{375} \;=\; \frac{1}{5}.$$

Substitute $$r = \frac{1}{5}$$ back into $$a = 15(1 - r)$$ to find $$a$$:

$$a \;=\; 15\Bigl(1 - \frac{1}{5}\Bigr) \;=\; 15\Bigl(\frac{4}{5}\Bigr) \;=\; 12.$$

The series whose sum we are now asked to evaluate is

$$ar^2 + ar^4 + ar^6 + \ldots$$

Its first term is

$$ar^2 \;=\; 12\Bigl(\frac{1}{5}\Bigr)^2 \;=\; 12 \times \frac{1}{25} \;=\; \frac{12}{25},$$

and its common ratio is

$$r^2 \;=\; \Bigl(\frac{1}{5}\Bigr)^2 \;=\; \frac{1}{25}.$$

Applying the infinite-GP sum formula one last time, we obtain

$$\text{Required sum} \;=\; \frac{\dfrac{12}{25}}{1 - \dfrac{1}{25}} \;=\; \frac{\dfrac{12}{25}}{\dfrac{24}{25}} \;=\; \frac{12}{25}\times\frac{25}{24} \;=\; \frac{12}{24} \;=\; \frac{1}{2}.$$

Hence, the correct answer is Option C.