Question

Determine the common ratio and find the next three terms of the geometric sequence 9,3sqrt3,3

Answer 1

Answer:

Fourth term: $a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{3}$ = $\sqrt{3}$

Fifth term: $a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{4}$ = 1

Sixth term: $a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}$

Step-by-step explanation:

The geometric progression is:

$9, 3 \sqrt{3}, 3...$

The first term, a, is 9

To find the common ratio, r, all we have to do is divide a term by its preceding term.

Let us divide the second term by the first:

$r = \frac{3\sqrt{3}}{9}\\ \\r = \frac{\sqrt{3}}{3}$

That is the common ratio.

Geometric progression is given generally as:

$a_n = ar^{(n - 1)}$

where a = first term

r = common ratio

$a_n$ = nth term

We need to find the 4th, 5th and 6th terms.

Fourth term: $a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{3}$ = $\sqrt{3}$

Fifth term: $a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{4}$ = 1

Sixth term: $a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}$