How many terms of the series,

$\dfrac{2}{9} - \dfrac{1}{3} + \dfrac{1}{2} + ….$ will make the sum $\dfrac{55}{72}$ ?

The above series is a G.P. with a = $\dfrac{2}{9} \text{ and } r = \dfrac{-\dfrac{1}{3}}{\dfrac{2}{9}} = -\dfrac{1}{3} \times \dfrac{9}{2} = -\dfrac{3}{2}$.

Let n terms be required to give the sum of $\dfrac{55}{72}$.

By formula,

$S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{-\dfrac{3}{2} - 1} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{2\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{9 \times \Big(\dfrac{-3 - 2}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{4\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{9 \times (-5)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{4\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{-45} \\[1em] \Rightarrow \dfrac{55 \times -45}{72 \times 4} = \Big(-\dfrac{3}{2}\Big)^n - 1 \\[1em] \Rightarrow -\dfrac{2475}{288} = \Big(-\dfrac{3}{2}\Big)^n - 1 \\[1em] \Rightarrow -\dfrac{2475}{288} + 1 = \Big(-\dfrac{3}{2}\Big)^n \\[1em] \Rightarrow \dfrac{-2475 + 288}{288} = \Big(-\dfrac{3}{2}\Big)^n \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{2187}{288} \\[1em]$

Dividing the numerator and denominator by 9

$\Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{243}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \Big(-\dfrac{3}{2}\Big)^5 \\[1em] \therefore n = 5.$

Hence, the required number of terms of the G.P. are 5.