The sum of a G.P. with common ratio r = 3 is | vedclass

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(A) The sum of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given $S_n = 364$,$r = 3$,and the last term $l = a r^{n-1} = 243$.We can rewrite

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$6$

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(A) The sum of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given $S_n = 364$,$r = 3$,and the last term $l = a r^{n-1} = 243$.We can rewrite the sum formula as $S_n = \frac{a r^{n-1} \cdot r - a}{r - 1}$.Substituting the known values: $364 = \frac{243 \cdot 3 - a}{3 - 1}$.$364 = \frac{729 - a}{2}$.$728 = 729 - a$,which gives $a = 1$.Now,use the last term formula: $a r^{n-1} = 243$.$1 \cdot 3^{n-1} = 243$.$3^{n-1} = 3^5$.Therefore,$n - 1 = 5$,which means $n = 6$.

The sum of a $G.P.$ with common ratio $r = 3$ is $364$,and the last term is $243$. Find the number of terms $n$.

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