Question

Find the mean of first n natural numbers.

Answer 1

Hint: We will first write the formula of the mean. Now, to put in the values, we will require to find the number of values and the sum of all the observations involved. Now, we will use the formula of sum of first n natural numbers and put in the formula and thus have the answer.

Complete step-by-step answer:
We know that the formula of mean is given by the following expression:-
Mean is equal to Sum of all the observations divided by the number of observations.
Now, we are given that we have to find the mean of first n natural numbers.
$\therefore $ we have a number of observations as n.
Now, we need to find the sum of all those observations.
The first n natural numbers are given by 1, 2, 3, ………, n.
We know that we have a formula of sum of first n natural numbers which is given by:-
Sum of first n natural numbers = $\sum {n = } \dfrac{{n(n + 1)}}{2}$.
Now, let us put these things in the formula $Mean = \dfrac{{Sum}}{{Frequency}}$:
$ \Rightarrow Mean = \dfrac{{\dfrac{{n(n + 1)}}{2}}}{n}$
Rewriting the RHS as follows:-
$ \Rightarrow Mean = \dfrac{{n(n + 1)}}{{2n}}$
Since, we know that n is not equal to 0. Therefore, we will get:-
$\therefore Mean = \dfrac{{n + 1}}{2}$

Hence, the answer is $\dfrac{{n + 1}}{2}$.

Note: The students must not be eager to learn the formula for first n natural numbers because it is damn easy to derive as well and that will help them build their brains as well. So, let us learn how to derive it:-
We know that the first n natural numbers are given by 1, 2, 3, ……., n.
Now, we see that we can call this sequence an arithmetic progression with a common difference of 2 – 1 = 1. So, we have an A. P. with a = 1 and d = 1.
We know that sum of n terms in an A. P. is given by ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
Now putting in the values, we will get:-
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2 \times 1 + (n - 1)1} \right]$
Now, simplifying the RHS, we will get:-
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {n + 1} \right]$
We can see that we have now derived it.