The sum of the 3$$^{rd}$$ and the 4$$^{th}$$ terms of a G.P. is 60 and the product of its first three terms is 1000. If the first term of this G.P. is positive, then its 7$$^{th}$$ term is:

Let us denote the first term of the geometric progression (G.P.) by $$a$$ and its common ratio by $$r$$.

For a G.P., the $$n^{\text{th}}$$ term is given by the well-known formula $$T_n = a\,r^{\,n-1}$$. Hence

$$T_3 = a\,r^{\,3-1}=a\,r^2 \quad\text{and}\quad T_4 = a\,r^{\,4-1}=a\,r^3.$$

According to the first piece of information, the sum of the third and the fourth terms equals 60:

$$T_3 + T_4 = 60 \;\Longrightarrow\; a\,r^2 + a\,r^3 = 60.$$

We can factor out the common part $$a\,r^2$$ from the left-hand side:

$$a\,r^2\,(1 + r) = 60. \cdots(1)$$

Next, we are told that the product of the first three terms is 1000. The first three terms are $$a,\; a\,r,\; a\,r^2,$$ so their product is

$$a \times a\,r \times a\,r^2 = a^3\,r^3.$$

Setting this equal to 1000 gives

$$a^3\,r^3 = 1000. \cdots(2)$$

From equation (2) we can take the real cube root on both sides. Because $$1000 = 10^3,$$ we obtain

$$a\,r = 10. \cdots(3)$$

Now we return to equation (1). First observe that

$$a\,r^2 = (a\,r)\,r.$$

Using the value from (3), namely $$a\,r = 10,$$ we substitute this into equation (1):

$$10\,r\,(1 + r) = 60.$$

Dividing every term by 10 simplifies the equation:

$$r\,(1 + r) = 6.$$

Expanding gives

$$r + r^2 = 6.$$

Re-writing in standard quadratic form,

$$r^2 + r - 6 = 0.$$

We factor this quadratic expression:

$$(r + 3)(r - 2) = 0.$$

Hence the possible values of the common ratio are

$$r = -3 \quad\text{or}\quad r = 2.$$

We must now examine the sign of $$a$$. From equation (3) we have $$a = \dfrac{10}{r}.$$

If $$r = -3,$$ then $$a = \dfrac{10}{-3} = -\dfrac{10}{3},$$ which is negative. Because the problem states that the first term is positive, this value of $$r$$ is not admissible.

Therefore we select

$$r = 2.$$

Substituting $$r = 2$$ into equation (3) immediately yields the first term:

$$a = \dfrac{10}{2} = 5.$$

We are now asked to find the seventh term $$T_7$$. Using the general term formula again,

$$T_7 = a\,r^{\,7-1} = a\,r^6.$$

With $$a = 5$$ and $$r = 2$$, we calculate

$$T_7 = 5 \times 2^6 = 5 \times 64 = 320.$$

Hence, the correct answer is Option A.