How many terms of the G.P. 3, frac{3}{2}, fra | vedclass

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(C) Let $n$ be the number of terms needed. Given that the first term $a = 3$ and the common ratio $r = \frac{1}{2}$.The sum of the first $n$ terms of

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$10$

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(C) Let $n$ be the number of terms needed. Given that the first term $a = 3$ and the common ratio $r = \frac{1}{2}$.The sum of the first $n$ terms of a $G.P.$ is given by $S_{n} = \frac{a(1 - r^{n})}{1 - r}$.Substituting the given values,we have $\frac{3069}{512} = \frac{3(1 - (\frac{1}{2})^{n})}{1 - \frac{1}{2}}$.$\frac{3069}{512} = \frac{3(1 - \frac{1}{2^{n}})}{\frac{1}{2}} = 6(1 - \frac{1}{2^{n}})$.Dividing both sides by $6$,we get $\frac{3069}{3072} = 1 - \frac{1}{2^{n}}$.$\frac{1}{2^{n}} = 1 - \frac{3069}{3072} = \frac{3072 - 3069}{3072} = \frac{3}{3072} = \frac{1}{1024}$.Since $1024 = 2^{10}$,we have $2^{n} = 2^{10}$,which implies $n = 10$.

How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$

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