If the sum of the second, fourth and sixth terms of a G.P.

Question

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 :

Accepted Answer

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 	ext{(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 $\frac{7}{91} 	imes (19683 - 1)/2 = \frac{7 	imes 19682}{91 	imes 2}$:

$ = \frac{7 	imes 19682}{91 	imes 2} = \frac{9841}{13} = 757 $

Therefore, the sum of the first nine terms of the G.P. is 757.

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