If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Let the first term of the G.P. be $$a$$ and the common ratio be $$r$$, where $$a \gt 0$$ and $$r \gt 0$$.

The $$n^{\text{th}}$$ term is $$T_n = a\,r^{\,n-1}$$.

The second, fourth and sixth terms are
$$T_2 = a\,r,$$ $$T_4 = a\,r^{3},$$ $$T_6 = a\,r^{5}.$$

Their sum is given to be $$21$$, so
$$a\,r + a\,r^{3} + a\,r^{5} = 21.$$

Factor out $$a\,r$$:
$$a\,r\bigl(1 + r^{2} + r^{4}\bigr) = 21 \qquad -(1)$$

The eighth, tenth and twelfth terms are
$$T_8 = a\,r^{7},$$ $$T_{10} = a\,r^{9},$$ $$T_{12} = a\,r^{11}.$$

Their sum is given to be $$15309$$, so
$$a\,r^{7} + a\,r^{9} + a\,r^{11} = 15309.$$

Factor out $$a\,r^{7}$$:
$$a\,r^{7}\bigl(1 + r^{2} + r^{4}\bigr) = 15309 \qquad -(2)$$

Both $$(1)$$ and $$(2)$$ contain the common factor $$\bigl(1 + r^{2} + r^{4}\bigr)$$. Divide $$(2)$$ by $$(1)$$:

$$\frac{a\,r^{7}\bigl(1 + r^{2} + r^{4}\bigr)}{a\,r\bigl(1 + r^{2} + r^{4}\bigr)} = \frac{15309}{21}$$
$$r^{6} = 729$$

Since $$729 = 3^{6}$$ and $$r \gt 0$$, we obtain
$$r = 3.$$

Compute $$1 + r^{2} + r^{4}$$:
$$1 + 3^{2} + 3^{4} = 1 + 9 + 81 = 91.$$

Substitute $$r = 3$$ into $$(1)$$ to find $$a$$:
$$a\,(3)\,(91) = 21$$
$$273\,a = 21$$
$$a = \frac{21}{273} = \frac{1}{13}.$$

The sum of the first nine terms of a G.P. with $$r \neq 1$$ is
$$S_9 = a\,\frac{r^{9} - 1}{r - 1}.$$

Substitute $$a = \frac{1}{13}$$ and $$r = 3$$:
$$S_9 = \frac{1}{13}\,\frac{3^{9} - 1}{3 - 1}.$$

Calculate $$3^{9}$$:
$$3^{9} = 19683.$$

Therefore
$$S_9 = \frac{1}{13}\,\frac{19683 - 1}{2} = \frac{1}{13}\,\frac{19682}{2} = \frac{19682}{26} = 757.$$

Hence, the sum of the first nine terms is $$757$$.

Option D is correct.