Find the sum of the series 81 - 27 + 9 - ….. - $\dfrac{1}{27}$.

The above series is in G.P. with a = 81, r = $-\dfrac{1}{3} \text{ and l } = -\dfrac{1}{27}.$

$\text{Sum } = \dfrac{a - lr}{1 - r} \\[1em] \therefore \text{Sum } = \dfrac{81 - \Big(-\dfrac{1}{27}\Big)(-\dfrac{1}{3})}{1 - \Big(-\dfrac{1}{3}\Big)} \\[1em] = \dfrac{81 - \dfrac{1}{81}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{6561 - 1}{81}}{\dfrac{3 + 1}{3}} \\[1em] = \dfrac{\dfrac{6560}{81}}{\dfrac{4}{3}} \\[1em] = \dfrac{6560 \times 3}{81 \times 4} \\[1em] = \dfrac{19680}{324}$

Dividing numerator and denominator by 12 we get,

$= \dfrac{1640}{27}.$

Hence, the sum of the series 81 - 27 + 9 - ….. - $\dfrac{1}{27} \text{ is } \dfrac{1640}{27}.$