Chapter 12.3, Problem 15AYU

To determine

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

Answer to Problem 15AYU

The sequence is geometric with common ratio $r=\frac{3}{2}$ and first four terms are . $\frac{1}{2},\frac{3}{4},\frac{9}{8}$ and $\frac{27}{16}$

.

Explanation of Solution

Given information:

The given relation is.

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

Calculation:

Consider the given relation.

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

Calculate first four term of the given sequence.

  $\begin{array}{c}{t}_1=\frac{3^{1-1}}{2^n}\\ =\frac{1}{2}\\ t_2=\frac{3^{2-1}}{2^2}\\ =\frac{3}{4}\end{array}$ .

  $\begin{array}{c}{t}_3=\frac{3^{3-1}}{2^3}\\ =\frac{9}{8}\\ t_4=\frac{3^{4-1}}{2^4}\\ =\frac{27}{16}\end{array}$

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

  $r=\frac{t_2}{t_1}=\frac{t_3}{t_2}$

Here $r$ is common ratio.

Put the values and simplify.

  $\begin{array}{c}r=\frac{\frac{3}{4}}{\frac{1}{2}}=\frac{\frac{9}{8}}{\frac{3}{4}}\\ r=\frac{3}{2}=\frac{3}{2}\end{array}$ .

Therefore, the sequence is geometric with common ratio $r=\frac{3}{2}$ and first four terms are . $\frac{1}{2},\frac{3}{4},\frac{9}{8}$ and $\frac{27}{16}$ .