How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$
How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$
Let $n$ be the number of terms needed. Given that $a=3, r=\frac{1}{2}$ and $S_{n}=\frac{3069}{512}$
Since $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
Therefore $\frac{3069}{512}=\frac{3\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^{n}}\right)$
or $\frac{3069}{3072}=1-\frac{1}{2^{n}}$
or $\frac{1}{2^{n}}=1-\frac{3069}{3072}=\frac{3}{3072}=\frac{1}{1024}$
or $2^{n}=1024=2^{10},$ which gives $n=10$
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