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 $S_6 = 9 	imes S_3$.Substituting the formula: $\frac{a(r^6 - 1)}{r - 1} = 9 	imes \frac{a(r^3 - 1)}{r - 1}$.Assuming $r 
eq 1$,we can cancel $\frac{a}{r - 1}$ from both sides:$r^6 - 1 = 9(r^3 - 1)$.Using the identity $x^2 - 1 = (x - 1)(x + 1)$ where $x = r^3$:$(r^3 - 1)(r^3 + 1) = 9(r^3 - 1)$.Since $r 
eq 1$,$r^3 - 1 
eq 0$,so we can divide by $(r^3 - 1)$:$r^3 + 1 = 9$.$r^3 = 8$.$r = 2$.

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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