Question

Show that the progression $$\frac{-3}{4}, \frac{1}{2}, \frac{-1}{3}, \frac{2}{9}$$ ......... is a G.P.

Write its (i) first term (ii) common ratio (iii) nth term (iv) 6th term.

Answer 1

To show it is a G.P.

$$\frac{\frac{1}{2}}{\frac{-3}{4}} = -\frac{2}{3}, \frac{\frac{-1}{3}}{\frac{1}{2}} = -\frac{2}{3}, \frac{\frac{2}{9}}{\frac{-1}{3}} = -\frac{2}{3}$$

All the ratios are the same (constant = $$-\frac{2}{3}$$), so the given numbers form a G.P.

(i) First term: $$a = -\frac{3}{4}$$

(ii) Common ratio: $$r = -\frac{2}{3}$$

(iii) nth term: $$T_n = ar^{n - 1} = -\frac{3}{4} \times \left(-\frac{2}{3}\right)^{n-1}$$

(iv) 6th term: $$T_6 = -\frac{3}{4} \times \left(-\frac{2}{3}\right)^{5}$$

$$= -\frac{3}{4} \times \left(-\frac{32}{243}\right)$$

$$= \frac{96}{972}$$

$$= \frac{8}{81}$$

So, the 6th term is $$\frac{8}{81}$$.