JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 11)
Explanation
To solve this problem, we have to work with two equations derived from the geometric progression (G.P.):
$\mathrm{ar} + \mathrm{ar}^3 + \mathrm{ar}^5 = 21$
$\mathrm{ar}^7 + \mathrm{ar}^9 + \mathrm{ar}^{11} = 15309$
From these equations, we can extract the terms:
Equation (1): $\mathrm{ar}(1 + \mathrm{r}^2 + \mathrm{r}^4) = 21$
Equation (2): $\mathrm{ar}^7(1 + \mathrm{r}^2 + \mathrm{r}^4) = 15309$
Now, divide equation (2) by equation (1):
$ \frac{\mathrm{ar}^7}{\mathrm{ar}} = \frac{15309}{21} $
From this division, we get:
$ \mathrm{r}^6 = 729 $
Which implies:
$ \mathrm{r} = 3 \quad \text{(since both terms and ratios are positive)} $
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.:
$ S_n = \frac{\mathrm{a}(\mathrm{r}^9 - 1)}{\mathrm{r} - 1} $
Substitute known values:
$ \Rightarrow \frac{\mathrm{a}(3^9 - 1)}{3 - 1} = \frac{\mathrm{a} \cdot (19683 - 1)}{2} $
Given that $\frac{7}{91} \times (19683 - 1)/2 = \frac{7 \times 19682}{91 \times 2}$:
$ = \frac{7 \times 19682}{91 \times 2} = \frac{9841}{13} = 757 $
Therefore, the sum of the first nine terms of the G.P. is 757.
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