Let S$$_n$$ denote the sum of the first n terms of an A.P. If S$$_4$$ = 16 and S$$_6$$ = -48, then S$$_{10}$$ is equal to:

We consider an arithmetic progression whose first term is denoted by $$a$$ and common difference by $$d$$.

For any arithmetic progression, the formula for the sum of the first $$n$$ terms is stated as

$$S_n=\dfrac{n}{2}\,\bigl[\,2a+(n-1)d\,\bigr].$$

We are told that the sum of the first four terms is $$S_4=16$$. Substituting $$n=4$$ into the sum formula, we obtain

$$S_4=\dfrac{4}{2}\,\bigl[\,2a+(4-1)d\,\bigr] \;=\;2\bigl[\,2a+3d\,\bigr].$$

Given that this value equals $$16$$, we write

$$2\bigl[\,2a+3d\,\bigr]=16.$$

Dividing both sides by $$2$$ yields

$$2a+3d=8.\qquad(1)$$

Next, the sum of the first six terms is given as $$S_6=-48$$. Putting $$n=6$$ into the same formula, we get

$$S_6=\dfrac{6}{2}\,\bigl[\,2a+(6-1)d\,\bigr] \;=\;3\bigl[\,2a+5d\,\bigr].$$

Equating this to $$-48$$ gives

$$3\bigl[\,2a+5d\,\bigr]=-48.$$

Dividing both sides by $$3$$ results in

$$2a+5d=-16.\qquad(2)$$

We now possess two linear equations, namely (1) $$2a+3d=8$$ and (2) $$2a+5d=-16$$. Subtracting equation (1) from equation (2) term by term, we have

$$\bigl(2a+5d\bigr)-\bigl(2a+3d\bigr)=(-16)-8,$$

which simplifies to

$$2d=-24.$$

Dividing by $$2$$, we secure the common difference:

$$d=-12.$$

Substituting this value of $$d$$ back into equation (1), we obtain

$$2a+3(-12)=8.$$

This simplifies to

$$2a-36=8,$$

so

$$2a=44$$

and consequently

$$a=22.$$

With both $$a$$ and $$d$$ known, we proceed to find the sum of the first ten terms $$S_{10}$$. Applying the sum formula once more with $$n=10$$, we write

$$S_{10}=\dfrac{10}{2}\,\bigl[\,2a+(10-1)d\,\bigr]=5\bigl[\,2a+9d\,\bigr].$$

Substituting $$a=22$$ and $$d=-12$$, we get

$$S_{10}=5\bigl[\,2(22)+9(-12)\bigr]=5\bigl[\,44-108\bigr]=5(-64)=-320.$$

Hence, the correct answer is Option A.