Let $$S_n$$ be the sum of the first $$n$$ terms of an arithmetic progression. If $$S_{3n} = 3S_{2n}$$, then the value of $$\frac{S_{4n}}{S_{2n}}$$ is:

Let the first term of the arithmetic progression be $$a$$ and the common difference be $$d$$.

For an arithmetic progression, the sum of the first $$m$$ terms is given by the well-known formula $$S_m=\frac{m}{2}\,[\,2a+(m-1)d\,].$$

We are told that $$S_{3n}=3S_{2n}$$. Substituting the formula for each sum, we have $$\frac{3n}{2}\,[\,2a+(3n-1)d\,]=3\Bigl(n\,[\,2a+(2n-1)d\,]\Bigr).$$

Dividing both sides by $$n$$ gives $$\frac{3}{2}\,[\,2a+(3n-1)d\,]=3[\,2a+(2n-1)d\,].$$

Multiplying every term by $$2$$ to clear the denominator, we obtain $$3[\,2a+(3n-1)d\,]=6[\,2a+(2n-1)d\,].$$

Dividing by $$3$$ simplifies this to $$2a+(3n-1)d=2[\,2a+(2n-1)d\,]=4a+(4n-2)d.$$

Bringing all terms to one side, we get $$0=4a-2a+(4n-2)d-(3n-1)d=2a+\bigl[(4n-2)-(3n-1)\bigr]d.$$

Simplifying the bracket, $$4n-2-3n+1=n-1$$, so the relation between $$a$$ and $$d$$ becomes $$2a+(n-1)d=0 \;\;\Longrightarrow\;\; 2a=-(n-1)d \;\;\Longrightarrow\;\; a=-\frac{(n-1)d}{2}.$$

Next, we need $$\dfrac{S_{4n}}{S_{2n}}$$. Using the sum formula again,

$$S_{4n}=\frac{4n}{2}\,[\,2a+(4n-1)d\,]=2n\,[\,2a+(4n-1)d\,],$$ $$S_{2n}=n\,[\,2a+(2n-1)d\,].$$

Therefore, $$\frac{S_{4n}}{S_{2n}}=\frac{2n\,[\,2a+(4n-1)d\,]}{n\,[\,2a+(2n-1)d\,]}=2\,\frac{2a+(4n-1)d}{2a+(2n-1)d}.$$

We now substitute $$a=-\dfrac{(n-1)d}{2}$$. First, compute $$2a$$: $$2a=-\,(n-1)d.$$

Numerator: $$2a+(4n-1)d=-(n-1)d+(4n-1)d=\bigl[-(n-1)+(4n-1)\bigr]d=(3n)d.$$

Denominator: $$2a+(2n-1)d=-(n-1)d+(2n-1)d=\bigl[-(n-1)+(2n-1)\bigr]d=(n)d.$$

Substituting these back, $$\frac{S_{4n}}{S_{2n}}=2\,\frac{3n\,d}{n\,d}=2\cdot 3=6.$$

Hence, the correct answer is Option A.