How many terms of the geometric progression (G.P) $\;$ $3, \; \dfrac{3}{2}, \; \dfrac{3}{4}, \cdots$ $\;$ are needed to give the sum $\dfrac{186}{32}$?
First term of G.P $= a = 3$
Common ratio of G.P $= r = \dfrac{3 / 2}{3} = \dfrac{1}{2} < 1$
Let, number of terms of G.P needed to get the required sum $= n$
Sum of G.P $= S_n = \dfrac{186}{32}$
Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(1 - r^n\right)}{1 - r}, \;\; r < 1$
i.e. $\;$ $\dfrac{186}{32} = \dfrac{3 \left[1 - \left(\dfrac{1}{2}\right)^n\right]}{1 - \dfrac{1}{2}}$
i.e. $\;$ $\dfrac{62}{32} = \dfrac{1 - \left(\dfrac{1}{2}\right)^n}{\dfrac{1}{2}}$
i.e. $\;$ $\dfrac{31}{16} \times \dfrac{1}{2} = 1 - \left(\dfrac{1}{2}\right)^n$
i.e. $\;$ $\left(\dfrac{1}{2}\right)^n = 1 - \dfrac{31}{32} = \dfrac{1}{32} = \left(\dfrac{1}{2}\right)^5$
$\implies$ $n = 5$