Chapter 12.3, Problem 21E

To determine

Calculate the first five terms of the given sequence $a_n=2{\left(3\right)}^n$ and find whether it is geometric. If it is geometric then also calculate the common ratio, and express the $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ .

Answer to Problem 21E

The first five term of the given sequence $a_n=2{\left(3\right)}^n$ is $6,18,54,162\&486$

The given sequence in geometric

Having the common ratio, $r=3$

The $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ is $a_n=6*3^{n-1}$

Explanation of Solution

Given information:

It is given in the question that,

the given sequence is $a_n=2{\left(3\right)}^n$

Formula used:

  $a_n=a{r}^{n-1}$ & $r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}$

Where,

  $\begin{array}{l}aisthe{1}^{st}term\\ nisthenumberofterm\\ risthecommonratio\end{array}$

Calculation:

For finding the first five term put $n=1,2,3,4,5$ in the given sequence is $a_n=2{\left(3\right)}^n$

So,

By putting $n=1$ in $a_n=2{\left(3\right)}^n$

  $\begin{array}{l}$ a_1=2{(3)}^1$=2*3\\ =6\end{array}$

By putting $n=2$ in $a_n=2{\left(3\right)}^n$

  $\begin{array}{l}$ a_2=2{(3)}^2$=2*9\\ =18\end{array}$

By putting $n=3$ in $a_n=2{\left(3\right)}^n$

  $\begin{array}{l}$ a_3=2{(3)}^3$=2*27\\ =54\end{array}$

By putting $n=4$ in $a_n=2{\left(3\right)}^n$

  $\begin{array}{l}$ a_4=2{(3)}^4$=2*81\\ =162\end{array}$

By putting $n=5$ in $a_n=2{\left(3\right)}^n$

  $\begin{array}{l}$ a_5=2{(3)}^5$=2*243\\ =486\end{array}$

  $\begin{array}{l}$ a_5=2{(3)}^5$=2*243\\ =486\end{array}$

By using the formula $r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}$

  $\begin{array}{l}r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}\\ \Rightarrow \frac{18}{6}=\frac{54}{18}\\ \Rightarrow 3=3\end{array}$

Here, the sequence satisfied the given formula concept so the given sequence $a_n=2{\left(3\right)}^n$ which shows that the common ratio of two consecutive term is same.

So, the given sequence in geometric

Having the common ratio, $r=3$

The $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ is $a_n=6*3^{n-1}$