Question

Find the sum of first 9 terms of the G.P. $$1, \frac{-1}{2}, \frac{1}{4}, \frac{-1}{8}, .........$$

Answer 1

Here, $$a = 1$$, $$r = -\dfrac{1}{2}$$ and $$n = 9$$.

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

$$S_9 = \frac{1 \times \left[1 - \left(-\frac{1}{2}\right)^9\right]}{\left(1 + \frac{1}{2}\right)}$$

$$= \frac{1 + \frac{1}{512}}{\frac{3}{2}}$$

$$= \frac{513}{512} \times \frac{2}{3}$$

$$= \frac{1026}{1536}$$

$$= \frac{171}{256}$$

Hence, the required sum is $$\dfrac{171}{256}$$.