Question

Which term of sequence $\sqrt{3}, 3, 3\sqrt{3}, \dots \dots \dots \dots \dots \dots \dots$ is 243?

Answer 1

The 10th term is $243$.

Answer 2

Given the sequence:

$$\sqrt{3},\, 3,\, 3\sqrt{3},\, \dots$$

This is a geometric progression.

• First term, $a = \sqrt{3}$
• Common ratio, $r = \frac{3}{\sqrt{3}} = \sqrt{3}$

The nth term is:

$$a_n = a \cdot r^{n-1} = \sqrt{3} \cdot (\sqrt{3})^{n-1} = (\sqrt{3})^n$$

We require:

$$(\sqrt{3})^n = 243$$

Expressing in terms of powers of 3:

$$(\sqrt{3})^n = 3^{\frac{n}{2}}$$

And since $243 = 3^5$, we equate:

$$3^{\frac{n}{2}} = 3^5 \implies \frac{n}{2} = 5 \implies n = 10$$