Question

Which term of the G.P. $$\sqrt{3}, 3, 3\sqrt{3}, 9$$, ..... is 729?

Answer 1

Here the first term $$a = \sqrt{3}$$ and the common ratio $$r = \frac{3}{\sqrt{3}} = \sqrt{3}$$.

So the nth term is

$$T_n = ar^{n-1}$$

$$= \sqrt{3} \times (\sqrt{3})^{n - 1}$$

$$= (\sqrt{3})^{n}$$

$$= 3^{\frac{n}{2}}$$

Let the 729 be the nth term:

$$3^{\frac{n}{2}} = 729 = 3^6$$

$$\frac{n}{2} = 6 \Rightarrow n = 12$$

Hence, 729 is the 12th term of the given G.P.