Chapter 12.3, Problem 13AYU

To determine

To find: Prove the sequence geometric and find common ratio along with first four terms.

Answer to Problem 13AYU

The sequence is geometric with common ratio $r=2^{\frac{1}{3}}$ and first four terms are . $2^{\frac{1}{3}},2^{\frac{2}{3}},2$ and $2^{\frac{4}{3}}$ .

Explanation of Solution

Given information:

The given relation is.

  $\left\{e_n\right\}=\left\{2^{\frac{n}{3}}\right\}$ .

Calculation:

Consider the given relation.

  $\left\{e_n\right\}=\left\{2^{\frac{n}{3}}\right\}$ .

Calculate first four term of the given sequence.

  $\begin{array}{c}{e}_1=2^{\frac{1}{3}}\\ e_2=2^{\frac{2}{3}}\end{array}$ .

  $\begin{array}{c}{e}_3=2^{\frac{3}{3}}\\ =2\\ e_4=2^{\frac{4}{3}}\end{array}$

For the sequence to be geometric this relation must be true.

  $r=\frac{e_2}{e_1}=\frac{e_3}{e_2}$

Here $r$ is common ratio.

Put the values and simplify.

  $\begin{array}{c}r=\frac{2^{\frac{2}{3}}}{2^{\frac{1}{3}}}=\frac{2}{2^{\frac{2}{3}}}\\ r=2^{\frac{1}{3}}=2^{\frac{1}{3}}\end{array}$ .

Therefore, the sequence is geometric with common ratio $r=2^{\frac{1}{3}}$ and first four terms are . $2^{\frac{1}{3}},2^{\frac{2}{3}},2$ and $2^{\frac{4}{3}}$ .