How many terms of the G.P. $$ 3, \dfrac{3}{2},\dfrac{3}{4}...$$ are needed to give the sum $$ \dfrac{3069}{512}?$$
Correct option is A. 10
Let the sum of n terms of the G.P. $$3, \dfrac{3}{2},\dfrac{3}{4},...$$ be $$ \dfrac{3069}{512}.$$ Then,
$$ 3\left \{ \dfrac{1-\left ( \dfrac{1}{2} \right )^{n}}{1-\dfrac{1}{2}} \right \} = \dfrac{3069}{512} \Rightarrow 1-\dfrac{1}{2^n}= \dfrac{1023}{1024} \Rightarrow \dfrac{1}{2^n} = \dfrac{1}{2^{10}} \Rightarrow n = 10 $$
Hence, the sum of 10 terms of the given G.P. is $$ \dfrac{3069}{512}.$$