Question

Find the G.P. whose 4th and 7th terms are $$\frac{1}{18}$$ and $$\frac{-1}{486}$$ respectively.

Answer 1

Let the first term be a and the common ratio be r.

$$T_4 = ar^3 = \frac{1}{18} \quad ...(i)$$

$$T_7 = ar^6 = -\frac{1}{486} \quad ...(ii)$$

Dividing (ii) by (i):

$$r^3 = \frac{-1}{486} \times 18$$

$$= -\frac{18}{486}$$

$$= -\frac{1}{27}$$

$$= \left(-\frac{1}{3}\right)^3$$ 

$$\Rightarrow r = -\frac{1}{3}$$

Putting $$r = -\frac{1}{3}$$ in (i):

$$a \times \left(-\frac{1}{3}\right)^3 = \frac{1}{18}$$

$$\Rightarrow a \times \left(-\frac{1}{27}\right) = \frac{1}{18}$$

$$\Rightarrow a = -\frac{27}{18} = -\frac{3}{2}$$

So the G.P. is $${-\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, -\frac{1}{54}, \frac{1}{162}, -\frac{1}{486}, ...}$$ with $$a = -\frac{3}{2}$$ and $$r = -\frac{1}{3}$$.