How many terms of the G.P. $\dfrac{2}{9}$, $-\dfrac{1}{3}$, $\dfrac{1}{2}$, ……… must be taken to make the sum equal to $\dfrac{55}{72}$?

In the given G.P.,

a = $\dfrac{2}{9}$

r = $\dfrac{\dfrac{-1}{3}}{\dfrac{2}{9}} = \dfrac{-9}{2 \times 3} = \dfrac{-3}{2}$.

Let sum of n terms be equal to $\dfrac{55}{72}$.

S_n = $\dfrac{55}{72}$.

We know that,

The sum of the first n terms of a G.P. is given by :

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big]}{1 - \Big(-\dfrac{3}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big]}{\Big(1 + \dfrac{3}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big]}{\Big(\dfrac{5}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} \times \Big(\dfrac{5}{2}\Big)= \dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big] \\[1em] \Rightarrow \dfrac{275}{144} = \dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big] \\[1em] \Rightarrow \dfrac{275}{144} \times \dfrac{9}{2} = \Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big] \\[1em] \Rightarrow \dfrac{275}{32} = 1 - \Big(-\dfrac{3}{2}\Big)^n \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = 1 - \dfrac{275}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \dfrac{32 - 275}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \dfrac{-243}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \Big(-\dfrac{3}{2}\Big)^5 \\[1em] \Rightarrow n = 5.$

Hence, n = 5.