If the sum of the second,fourth and sixth ter | vedclass

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(D) Let the $G.P.$ be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given $ar + ar^3 + ar^5 = 21 \Rightarrow ar(1 + r^2 + r^4) = 21$ $(1)$.Given $ar^

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$757$

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(D) Let the $G.P.$ be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given $ar + ar^3 + ar^5 = 21 \Rightarrow ar(1 + r^2 + r^4) = 21$ $(1)$.Given $ar^7 + ar^9 + ar^{11} = 15309 \Rightarrow ar^7(1 + r^2 + r^4) = 15309$ $(2)$.Dividing $(2)$ by $(1)$,we get $\frac{ar^7(1 + r^2 + r^4)}{ar(1 + r^2 + r^4)} = \frac{15309}{21}$.$r^6 = 729$ $\Rightarrow r^6 = 3^6$ $\Rightarrow r = 3$ (since terms are positive).Substitute $r = 3$ into $(1)$: $a(3)(1 + 9 + 81) = 21$ $\Rightarrow 3a(91) = 21$ $\Rightarrow a = \frac{21}{273} = \frac{1}{13}$.The sum of the first $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1}$.For $n = 9$,$S_9 = \frac{\frac{1}{13}(3^9 - 1)}{3 - 1} = \frac{19683 - 1}{13 	imes 2} = \frac{19682}{26} = 757$.

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:

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