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

Here, a = 3 and r = $\dfrac{\dfrac{3}{2}}{3} = \dfrac{3}{2 \times 3} = \dfrac{1}{2}.$

Let n terms are needed for the sum.

$S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \Rightarrow \dfrac{3069}{512} = \dfrac{3\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big]}{\dfrac{1}{2} - 1} \\[1em] \Rightarrow \dfrac{3069}{512} = \dfrac{3\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big]}{\dfrac{1 - 2}{2}} \\[1em] \Rightarrow \dfrac{3069}{512} = -6\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big] \\[1em] \Rightarrow \dfrac{3069}{512 \times 6} = 1 - \Big(\dfrac{1}{2}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = 1 - \dfrac{3069}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{3072 - 3069}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{3}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{1}{1024} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \Big(\dfrac{1}{2}\Big)^{10} \\[1em] \therefore n = 10.$

Hence, 10 terms of the G.P. are required for the sum of $\dfrac{3069}{512}$.