Chapter 12.3, Problem 23E

To determine

Calculate the first five terms of the given sequence $a_n=\frac{1}{4^n}$ 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 23E

The first five term of the given sequence $a_n=\frac{1}{4^n}$ is $\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256}\&\frac{1}{1024}$

The given sequence in geometric

Having the common ratio, $r=\frac{1}{4}$

The $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ is $a_n=\frac{1}{4}*{\left(\frac{1}{4}\right)}^{n-1}$

Explanation of Solution

Given information:

It is given in the question that,

the given sequence is $a_n=\frac{1}{4^n}$

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 $a_n=\frac{1}{4^n}$

So,

By putting $n=1$ in $a_n=\frac{1}{4^n}$

  $\begin{array}{l}{a}_1=\frac{1}{4^1}\\ a_1=\frac{1}{4}\end{array}$

By putting $n=2$ in $a_n=\frac{1}{4^n}$

  $\begin{array}{l}{a}_2=\frac{1}{4^2}\\ =\frac{1}{16}\end{array}$

By putting $n=3$ in $a_n=\frac{1}{4^n}$

  $\begin{array}{l}{a}_3=\frac{1}{4^3}\\ =\frac{1}{64}\end{array}$

By putting $n=4$ in $a_n=\frac{1}{4^n}$

  $\begin{array}{l}{a}_4=\frac{1}{4^4}\\ =\frac{1}{256}\end{array}$

By putting $n=5$ in $a_n=\frac{1}{4^n}$

  $\begin{array}{l}{a}_5=\frac{1}{4^5}\\ =\frac{1}{1024}\end{array}$

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

  $\begin{array}{l}r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}\\ \Rightarrow \frac{\frac{1}{16}}{\frac{1}{4}}=\frac{\frac{1}{64}}{\frac{1}{16}}\\ \Rightarrow \frac{1}{16}*\frac{4}{1}=\frac{1}{64}*\frac{16}{1}\\ \Rightarrow \frac{1}{4}=\frac{1}{4}\end{array}$

Here, the sequence satisfied the given formula concept so the given sequence $a_n=\frac{1}{4^n}$ which shows that the common ratio of two consecutive term is same.

So, the given sequence in geometric

Having the common ratio, $r=\frac{1}{4}$

The $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ is $a_n=\frac{1}{4}*{\left(\frac{1}{4}\right)}^{n-1}$