Chapter 9, Problem 1RE

To determine

The first five terms of the sequence $a_n=\frac{n}{n+1}$ and state whether the given sequence is arithmetic sequence, geometric sequence or neither.

Answer to Problem 1RE

The first five terms of the sequence $a_n=\frac{n}{n+1}$ are $\underset{\_}{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}}$ and $\underset{\_}{\frac{5}{6}}$, the given sequence is neither arithmetic sequence nor geometric sequence.

Explanation of Solution

Given:

The sequence is $a_n=\frac{n}{n+1}$.

Definition used:

Arithmetic sequence:

A sequence in which the difference between the two successive terms is always constant is called an arithmetic sequence.

Geometric sequence:

A sequence in which the ratio between the successive terms is constant is called a geometric sequence.

Calculation:

To calculate the first five terms of the sequence $a_n=\frac{n}{n+1}$, substitute the value of $n$ as $n=1,2,3,4\text{and 5}$.

For $n=1$:

$\begin{array}{c}{a}_1=\frac{1}{1+1}\\ =\frac{1}{2}\end{array}$

For $n=2$:

$\begin{array}{c}{a}_2=\frac{2}{2+1}\\ =\frac{2}{3}\end{array}$

For $n=3$:

$\begin{array}{c}{a}_3=\frac{3}{3+1}\\ =\frac{3}{4}\end{array}$

For $n=4$:

$\begin{array}{c}{a}_4=\frac{4}{4+1}\\ =\frac{4}{5}\end{array}$

For $n=5$:

$\begin{array}{c}{a}_5=\frac{5}{5+1}\\ =\frac{5}{6}\end{array}$

Check whether the sequence is an arithmetic sequence.

A sequence in which the difference between the two successive terms is always constant is called an arithmetic sequence.

Difference between $a_1$ and $a_2$ is calculated as follows:

$\begin{array}{c}{a}_1-a_2=\frac{1}{2}-\frac{2}{3}\left(a_1=\frac{1}{2},a_2=\frac{2}{3}\right)\\ =\frac{3-4}{6}\\ =\frac{-1}{6}\end{array}$

Difference between $a_2$ and $a_3$ is calculated as follows:

$\begin{array}{c}{a}_2-a_3=\frac{2}{3}-\frac{3}{4}\left(a_2=\frac{2}{3},a_3=\frac{3}{4}\right)\\ =\frac{8-9}{12}\\ =\frac{-1}{12}\end{array}$

The difference between the terms $a_1,a_2$ is not equal to the difference between the terms $a_2,a_3$.

This implies that the sequence $a_n=\frac{n}{n+1}$ is not an arithmetic sequence.

Check whether the sequence is a geometric sequence.

A sequence in which the ratio between the successive terms is constant is called a geometric sequence.

Ratio between the terms $a_1$ and $a_2$ is calculated as follows:

$\begin{array}{c}\frac{a_1}{a_2}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{2}{3}\right)}\left(a_1=\frac{1}{2},a_2=\frac{2}{3}\right)\\ =\frac{1}{2}÷\frac{2}{3}\\ =\frac{1}{2}\times \frac{3}{2}\left(\frac{\text{a}}{\text{b}}÷\frac{\text{c}}{\text{d}}=\frac{\text{a}}{\text{b}}\times \frac{\text{d}}{\text{c}}\right)\\ =\frac{3}{4}\end{array}$

Ratio between the terms $a_2$ and $a_3$ is calculated as follows:

$\begin{array}{c}\frac{a_2}{a_3}=\frac{\left(\frac{2}{3}\right)}{\left(\frac{3}{4}\right)}\left(a_2=\frac{2}{3},a_3=\frac{3}{4}\right)\\ =\frac{2}{3}÷\frac{3}{4}\\ =\frac{2}{3}\times \frac{4}{3}\left(\frac{\text{a}}{\text{b}}÷\frac{\text{c}}{\text{d}}=\frac{\text{a}}{\text{b}}\times \frac{\text{d}}{\text{c}}\right)\\ =\frac{8}{9}\end{array}$

The ratio between the terms $a_1,a_2$ is not equal to the ratio between the terms $a_2,a_3$.

This implies that the sequence $a_n=\frac{n}{n+1}$ is not a geometric sequence.

Thus, the first five terms of the sequence $a_n=\frac{n}{n+1}$ are $\underset{\_}{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}}$ and $\underset{\_}{\frac{5}{6}}$, the given sequence is neither arithmetic sequence nor geometric sequence.