The sum of a few terms of a geometric series | vedclass

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Question

(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.Given the $n^{th}$ term $l = a r^{n-1} = 486$.So,$a(3)^{n-1} = 486$,

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$2$

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(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.Given the $n^{th}$ term $l = a r^{n-1} = 486$.So,$a(3)^{n-1} = 486$,which implies $a \cdot \frac{3^n}{3} = 486$,or $a \cdot 3^n = 1458$ $(i)$.The sum of $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1} = 728$.Substituting $r = 3$: $\frac{a(3^n - 1)}{3 - 1} = 728$.$a(3^n - 1) = 728 	imes 2 = 1456$.$a \cdot 3^n - a = 1456$ $(ii)$.Substituting $(i)$ into $(ii)$: $1458 - a = 1456$.$a = 1458 - 1456 = 2$.

The sum of a few terms of a geometric series is $728$. If the common ratio is $3$ and the last term is $486$,then the first term of the series will be:

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