In a geometric progression with n terms and a | vedclass

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(C) The sum of $n$ terms of a geometric progression is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.We know the last term $l = ar^{n-1} = 243$ and $r = 3

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$6$

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(C) The sum of $n$ terms of a geometric progression is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.We know the last term $l = ar^{n-1} = 243$ and $r = 3$.Substituting $ar^n = l 	imes r = 243 	imes 3 = 729$ into the sum formula:$S_n = \frac{ar^n - a}{r - 1} = 364$$\frac{729 - a}{3 - 1} = 364$$729 - a = 364 	imes 2$$729 - a = 728$$a = 1$Now,using $ar^{n-1} = 243$ with $a = 1$ and $r = 3$:$1 	imes 3^{n-1} = 243$$3^{n-1} = 3^5$$n - 1 = 5$$n = 6$.

In a geometric progression with $n$ terms and a common ratio of $3$,the sum of the $n$ terms is $364$ and the last term is $243$. Find the value of $n$.

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