The sum of the first four terms of a G.P. is | vedclass

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(C) Let the first four terms of the $G.P.$ be $a, ar, ar^2, ar^3$. Given,common ratio $r = 3$ and the sum of the first four terms $S_4 = 160$. The for

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$108$

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(C) Let the first four terms of the $G.P.$ be $a, ar, ar^2, ar^3$. Given,common ratio $r = 3$ and the sum of the first four terms $S_4 = 160$. The formula for the sum of the first $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1}$. Substituting the values: $160 = \frac{a(3^4 - 1)}{3 - 1}$. $160 = \frac{a(81 - 1)}{2} = \frac{a(80)}{2} = 40a$. Thus,$a = \frac{160}{40} = 4$. The $4^{th}$ term is $T_4 = ar^3$. $T_4 = 4 	imes (3)^3 = 4 	imes 27 = 108$.

The sum of the first four terms of a $G.P.$ is $160$ and the common ratio is $3$. Find the $4^{th}$ term.

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