Question

Find the 6th term from the end of the G.P. $$16, 8, 4, 2, ......., \frac{1}{512}$$. 

Answer 1

Here the first term $$a = 16$$, the common ratio $$r = \frac{8}{16} = \frac{1}{2}$$ and the last term $$l = \frac{1}{512}$$.

The nth term from the end is $$\frac{l}{r^{(n - 1)}}$$.

So the 6th term from the end is

$$\frac{l}{r^{(6 - 1)}} = \frac{\frac{1}{512}}{\left(\frac{1}{2}\right)^5}$$

$$= \frac{\frac{1}{512}}{\frac{1}{32}}$$

$$= \frac{32}{512}$$

$$= \frac{1}{16}$$

So, the 6th term from the end is $$\frac{1}{16}$$.