Question

Find the sum of first 6 terms of the G.P. 0.1, 0.01, 0.001, .........

Answer 1

Here, $$a = 0.1$$, $$r = \dfrac{0.01}{0.1} = 0.1$$ and $$n = 6$$.

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

$$S_6 = \frac{0.1 \times (1 - 0.1^6)}{(1 - 0.1)}$$

$$= \frac{0.1 \times \left(1 - \frac{1}{1000000}\right)}{0.9}$$

$$= \frac{1}{9} \times \frac{999999}{1000000}$$

$$= \frac{111111}{1000000}$$

$$= 0.111111$$

Hence, the required sum is 0.111111 (exactly $$\dfrac{111111}{1000000}$$).