Question

Show that the progression $$625, 125, 25, 5, 1, \frac{1}{5}$$, ......... is a G.P.

Write its (i) first term (ii) common ratio (iii) nth term (iv) 10th term.

Answer 1

To show it is a G.P.

$$\frac{125}{625} = \frac{1}{5}, \frac{25}{125} = \frac{1}{5}, \frac{5}{25} = \frac{1}{5}, \frac{1}{5}{1} = \frac{1}{5}, \frac{\frac{1}{5}}{\frac{1}{5}} = \frac{1}{5}$$

All the ratios are the same (constant = $$\frac{1}{5}$$), so the given numbers form a G.P.

(i) First term: $$a = 625$$

(ii) Common ratio: $$r = \frac{1}{5}$$

(iii) nth term: $$T_n = ar^{n - 1}$$ 

$$= 625 \times \left(\frac{1}{5}\right)^{n - 1}$$

$$= \frac{625}{5^{n - 1}}$$

$$= 5^{5 - n}$$

(iv) 10th term: $$T_{10} = \frac{625}{5^{(10 - 1)}}$$

$$= \frac{625}{5^9}$$

$$= \frac{5^4}{5^9}$$

$$= \frac{1}{5^5}$$

$$= \frac{1}{3125}$$

So, the 10th term is $$\frac{1}{3125}$$.