Question

Find the sum of first 6 terms of the G.P. $$1, \frac{-1}{3}, \frac{1}{3^2}, \frac{-1}{3^3}, ..........$$

Answer 1

Here, $$a = 1$$, $$r = -\dfrac{1}{3}$$ and $$n = 6$$.

Since $$r < 1$$, use $$S_n = \dfrac{a(1 - r^n)}{1 - r}$$.

$$S_6 = \frac{1 \times \left[1 - \left(-\frac{1}{3}\right)^6\right]}{\left(1 + \frac{1}{3}\right)}$$

$$= \frac{1 - \frac{1}{729}}{\frac{4}{3}}$$

$$= \frac{728}{729} \times \frac{3}{4}$$

$$= \frac{182}{243}$$

Hence, the required sum is $$\dfrac{182}{243}$$.