Chapter 12.3, Problem 16AYU

To determine

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

Answer to Problem 16AYU

The sequence is geometric with common ratio $r=\frac{2}{3}$ and first four terms are . $2,\frac{4}{3},\frac{8}{9}$ and $\frac{16}{27}$ .

Explanation of Solution

Given information:

The given relation is.

  $\left\{u_n\right\}=\left\{\frac{2^n}{3^{n-1}}\right\}$ .

Calculation:

Consider the given relation.

  $\left\{u_n\right\}=\left\{\frac{2^n}{3^{n-1}}\right\}$ .

Calculate first four term of the given sequence.

  $\begin{array}{c}{u}_1=\frac{2^1}{3^{1-1}}\\ =2\\ u_2=\frac{2^2}{3^{2-1}}\\ =\frac{4}{3}\end{array}$ .

  $\begin{array}{c}{u}_3=\frac{2^3}{3^{3-1}}\\ =\frac{8}{9}\\ u_4=\frac{2^4}{3^{4-1}}\\ =\frac{16}{27}\end{array}$

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

  $r=\frac{u_2}{u_1}=\frac{u_3}{u_2}$

Here $r$ is common ratio.

Put the values and simplify.

  $\begin{array}{c}r=\frac{\frac{4}{3}}{2}=\frac{\frac{8}{9}}{\frac{4}{3}}\\ r=\frac{2}{3}=\frac{2}{3}\end{array}$ .

Therefore, the sequence is geometric with common ratio $r=\frac{2}{3}$ and first four terms are . $2,\frac{4}{3},\frac{8}{9}$ and $\frac{16}{27}$ .