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 :

<p>To solve this problem, we have to work with two equations derived from the geometric progression (G.P.):</p> <p><p>$\mathrm{ar} + \mathrm{ar}^3 + \mathrm{ar}^5 = 21$</p></p> <p><p>$\mathrm{ar}^7 + \mathrm{ar}^9 + \mathrm{ar}^{11} = 15309$</p></p> <p>From these equations, we can extract the terms:</p> <p><p>Equation (1): $\mathrm{ar}(1 + \mathrm{r}^2 + \mathrm{r}^4) = 21$</p></p> <p><p>Equation (2): $\mathrm{ar}^7(1 + \mathrm{r}^2 + \mathrm{r}^4) = 15309$</p></p> <p>Now, divide equation (2) by equation (1):</p> <p>$ \frac{\mathrm{ar}^7}{\mathrm{ar}} = \frac{15309}{21} $</p> <p>From this division, we get:</p> <p>$ \mathrm{r}^6 = 729 $</p> <p>Which implies:</p> <p>$ \mathrm{r} = 3 \quad \text{(since both terms and ratios are positive)} $</p> <p>Using this value of $\mathrm{r}$, calculate the sum of the first nine terms of the G.P. using the formula for the sum of a G.P.:</p> <p>$ S_n = \frac{\mathrm{a}(\mathrm{r}^9 - 1)}{\mathrm{r} - 1} $</p> <p>Substitute known values:</p> <p>$ \Rightarrow \frac{\mathrm{a}(3^9 - 1)}{3 - 1} = \frac{\mathrm{a} \cdot (19683 - 1)}{2} $</p> <p>Given that $\frac{7}{91} \times (19683 - 1)/2 = \frac{7 \times 19682}{91 \times 2}$:</p> <p>$ = \frac{7 \times 19682}{91 \times 2} = \frac{9841}{13} = 757 $</p> <p>Therefore, the sum of the first nine terms of the G.P. is <strong>757</strong>.</p>