Question

The 4th, 6th and last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.

Answer 1

Let the first term be a, the common ratio be r and the number of terms be n.

$$T_4 = ar^3 = 10 \quad ...(i)$$

$$T_6 = ar^5 = 40 \quad ...(ii)$$

Dividing (ii) by (i):

$$r^2 = \frac{40}{10} = 4$$

$$\Rightarrow r = 2 \text{ (since } r \text{ is positive)}$$

Putting r = 2 in (i):

$$a \times 2^3 = 10$$

$$\Rightarrow 8a = 10$$

$$\Rightarrow a = \frac{10}{8} = \frac{5}{4}$$

For the last term:

$$l = ar^{n - 1} \Rightarrow \frac{5}{4} \times 2^{n - 1} = 640$$

$$2^{n - 1} = 640 \times \frac{4}{5} = 512 = 2^9$$

$$\Rightarrow n - 1 = 9$$

$$\Rightarrow n = 10$$

So, the first term is $$\frac{5}{4}$$, the common ratio is 2 and the G.P. has 10 terms.