Chapter 9.3, Problem 12E

To determine

To find whether the sequence $2,\frac{4}{\sqrt{3}},\frac{8}{3},\frac{16}{3\sqrt{3}},...$ is geometric and if so, then find its common ratio.

Answer to Problem 12E

The sequence $2,\frac{4}{\sqrt{3}},\frac{8}{3},\frac{16}{3\sqrt{3}},...$ is geometric and the common ratio $r=\frac{2}{\sqrt{3}}$ .

Explanation of Solution

Given information :

Let the sequence be $2,\frac{4}{\sqrt{3}},\frac{8}{3},\frac{16}{3\sqrt{3}},...$ .

Concept used:

In a sequence $a_1,a_2,a_3,a_4,\dots \dots ,a_n,\dots$ if ratio of the consecutive terms is the same then it is defined as Geometric Sequence.

So, a number r is defined as the common ratio of the geometric sequence if $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\dots \dots =r.$

Calculations:

Here, $a_1=2,a_2=\frac{4}{\sqrt{3}},a_3=\frac{8}{3},a_4=\frac{16}{3\sqrt{3}},\dots$

Calculate the ratio of the consecutive terms:

  $\begin{array}{l}\frac{a_2}{a_1}=\frac{\left(\frac{4}{\sqrt{3}}\right)}{2}=\frac{2}{\sqrt{3}}\\ \frac{a_3}{a_2}=\frac{\left(\frac{8}{3}\right)}{\left(\frac{4}{\sqrt{3}}\right)}=\frac{8\times \sqrt{3}}{4\times 3}=\frac{2}{\sqrt{3}}\\ \frac{a_4}{a_3}=\frac{\left(\frac{16}{3\sqrt{3}}\right)}{\left(\frac{8}{3}\right)}=\frac{16\times 3}{8\times 3\sqrt{3}}=\frac{2}{\sqrt{3}}\\ ⋮\end{array}$

And so on.

Since, $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\dots \dots =r=\frac{2}{\sqrt{3}}$ , therefore the sequence is geometric.