Question

Find the arithmetic mean of the series \[1,3,5,...........\left( {2n - 1} \right)\]
A) $n$
B) $2n$
C) $\dfrac{n}{2}$
D) $n - 1$

Answer 1

Hint:
Here, we will find the arithmetic mean of the given series. We will use the arithmetic progression to find the number of terms in the series, and then we will use the sum of the first $n$ terms in the arithmetic series to find the sum of the series. Then by using the arithmetic mean formula we will find the arithmetic mean. Arithmetic mean is defined as the average of the given numbers.

Formula Used:
We will use the following formulas:
1) Arithmetic Progression is given by the formula ${S_n} = a + \left( {k - 1} \right)d$ where $a$ is the first term and $d$ is the common difference.
2) Sum of the first $n$terms in an arithmetic progress is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where $a$is the first term and $d$is the common difference.
3) Arithmetic mean is given by the formula ${\text{Mean}} = \dfrac{{\sum x }}{n}$ where $\sum x $ is the sum of the series and $n$is the total number of series.

Complete Step by Step Solution:
We are given a number series is \[1, 3, 5, ...........\left( {2n - 1} \right)\]
The given series is in the form of Arithmetic Progression as $a,a + d,a + 2d,...........$
Therefore, the first term of an Arithmetic Progression $a = 1$, common difference $d = {a_2} - {a_1} = 3 - 1 = 2$ .
Let $\left( {2n - 1} \right)$ be the ${k^{th}}$term.
By substituting $a = 1$ and the $d = 2$ in the formula ${S_n} = a + \left( {k - 1} \right)d$, we get
$\left( {2n - 1} \right) = 1 + \left( {k - 1} \right)2$
By multiplying the terms, we get
$ \Rightarrow \left( {2n - 1} \right) = 1 + 2k - 2$
Subtracting the like term, we get
$ \Rightarrow \left( {2n - 1} \right) = 2k - 1$
Adding 1 on both sides, we get
$ \Rightarrow 2n = 2k$
$ \Rightarrow n = k$
Thus, $\left( {2n - 1} \right)$ be the ${n^{th}}$ term.
By substituting $a = 1$ and the $d = 2$ in the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, we get
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2\left( 1 \right) + \left( {n - 1} \right)2} \right]$
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2 + \left( {n - 1} \right)2} \right]$
Multiplying the terms, we get
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]$
Subtracting like terms, we get
$ \Rightarrow {S_n} = \dfrac{n}{2} \cdot 2n$
Dividing the terms, we get
$ \Rightarrow {S_n} = n \cdot n$
$ \Rightarrow {S_n} = {n^2}$

We are given that the sum of the series is ${n^2}$ and the total number of series is $n$.
Using the formula ${\text{Mean}} = \dfrac{{\sum x }}{n}$, we get
Arithmetic mean $ = \dfrac{{{n^2}}}{n}$
$ \Rightarrow $ Arithmetic mean $ = n$
Therefore, the arithmetic mean of the series \[1,3,5,...........\left( {2n - 1} \right)\] is $n$.

Thus, option (A) is the correct answer.

Note:
We should know that we are not given any value for $n$. We know that it is impossible to find the arithmetic mean without calculating the sum of the terms and the number of terms. So, we will prove that the series is in arithmetic progression, then we would find the sum of the first $n$ terms by using the concept of the arithmetic progression. An arithmetic progression is a type of progression or series in which the consecutive terms have the same common difference.