How many terms of the G.P. 3, 3^{2}, 3^{3}, d | vedclass

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(C) The given $G.P.$ is $3, 3^{2}, 3^{3}, \dots$Let $n$ terms of this $G.P.$ be required to obtain the sum $120$.The formula for the sum of $n$ terms

Vedclass · Accepted Answer

$4$

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(C) The given $G.P.$ is $3, 3^{2}, 3^{3}, \dots$Let $n$ terms of this $G.P.$ be required to obtain the sum $120$.The formula for the sum of $n$ terms of a $G.P.$ is $S_{n} = \frac{a(r^{n}-1)}{r-1}$ for $r > 1$.Here,the first term $a = 3$ and the common ratio $r = \frac{3^{2}}{3} = 3$.Substituting the values into the formula:$120 = \frac{3(3^{n}-1)}{3-1}$$120 = \frac{3(3^{n}-1)}{2}$$120 	imes \frac{2}{3} = 3^{n}-1$$40 	imes 2 = 3^{n}-1$$80 = 3^{n}-1$$3^{n} = 81$$3^{n} = 3^{4}$Therefore,$n = 4$.Thus,$4$ terms of the given $G.P.$ are required to obtain the sum $120$.

How many terms of the $G.P.$ $3, 3^{2}, 3^{3}, \dots$ are needed to give the sum $120$?

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