Chapter 12.3, Problem 10AYU

To determine

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

Answer to Problem 10AYU

The sequence is geometric with common ratio $r=\frac{5}{2}$ and first four terms are . $\frac{5}{2},\frac{25}{4},\frac{125}{8}$ and $\frac{625}{16}$ .

Explanation of Solution

Given information:

The given relation is.

  $\left\{b_n\right\}=\left\{{\left(\frac{5}{2}\right)}^n\right\}$ .

Calculation:

Consider the given relation.

  $\left\{b_n\right\}=\left\{{\left(\frac{5}{2}\right)}^n\right\}$ .

Calculate first four term of the given sequence.

  $\begin{array}{c}{b}_1={\left(\frac{5}{2}\right)}^1\\ =\frac{5}{2}\\ b_2={\left(\frac{5}{2}\right)}^2\\ =\frac{25}{4}\end{array}$ .

  $\begin{array}{c}{b}_3={\left(\frac{5}{2}\right)}^3\\ =\frac{125}{8}\\ b_4={\left(\frac{5}{2}\right)}^4\\ =\frac{625}{16}\end{array}$

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

  $r=\frac{b_2}{b_1}=\frac{b_3}{b_2}$

Here $r$ is common ratio.

Put the values and simplify.

  $\begin{array}{c}r=\frac{\frac{25}{4}}{\frac{5}{2}}=\frac{\frac{125}{8}}{\frac{25}{4}}\\ r=\frac{5}{2}=\frac{5}{2}\end{array}$ .

Therefore, the sequence is geometric with common ratio $r=\frac{5}{2}$ and first four terms are . $\frac{5}{2},\frac{25}{4},\frac{125}{8}$ and $\frac{625}{16}$ .