If the sum of the first 6 terms is 9 times th | vedclass

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(B) Let the first term be $a$ and the common ratio be $r$. The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Giv

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$2$

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(B) Let the first term be $a$ and the common ratio be $r$. The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given that $S_6 = 9 	imes S_3$,we have:$\frac{a(r^6 - 1)}{r - 1} = 9 	imes \frac{a(r^3 - 1)}{r - 1}$Assuming $r 
eq 1$,we can simplify this to:$r^6 - 1 = 9(r^3 - 1)$$(r^3)^2 - 1 = 9(r^3 - 1)$Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get:$(r^3 - 1)(r^3 + 1) = 9(r^3 - 1)$Since $r 
eq 1$,$r^3 - 1 
eq 0$,so we can divide both sides by $(r^3 - 1)$:$r^3 + 1 = 9$$r^3 = 8$$r = 2$Thus,the common ratio is $2$.

If the sum of the first $6$ terms is $9$ times the sum of the first $3$ terms of the same $G.P.$,then the common ratio of the series will be

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