Question

The arithmetic mean of the first $n$ odd numbers is:
A. $n$
B. $\dfrac{n}{2}$
C. $\dfrac{{n - 1}}{2}$
D. $\dfrac{{n + 1}}{2}$

Answer 1

Hint:
Odd numbers starts from one and the common difference is $2$ which means that $1, 3, 5, 7, 9........$ are all the odd numbers and the arithmetic mean of the first $n$ odd numbers is $\dfrac{{{\text{sum of n numbers}}}}{n}$
Sum of the AP is given as $\dfrac{n}{2}(2a + (n - 1)d)$

Complete step by step solution:
Here in this question we are given that if we have the first n odd numbers then we need to find what their arithmetic mean will be? So here we need to understand the two things. Firstly the means basically mean is the average of the two or more terms like mean of the numbers $10{\text{ and 20}}$ is $15$ and here now we have first $n$ odd numbers whose mean we have to find.
Basically mean is denoted by $\overline x $ and its formula is given by
Hence $\overline x = \dfrac{{{x_1} + {x_2} + {x_3}........... + {x_n}}}{n}$ which is given as $\dfrac{{{\text{sum of n numbers}}}}{n}$
Here we understood what is the mean but now we need to know what is meant by an odd number. Odd numbers starts from one and the common difference is $2$ which means that $1,3,5,7,9........$ which means these are those numbers which leave the reminder one when divided by $2$
So according to the given $n$ terms we have the odd numbers as $1,3,5,7,9........{\text{till n terms}}$
So basically here the total number of terms$ = n$
Common difference$ = 3 - 1 = 5 - 3 = 2$
These are in AP and we know that
Sum of the AP is given as $\dfrac{n}{2}(2a + (n - 1)d)$
Sum will be given as
$
   = \dfrac{n}{2}(2(1) + (n - 1)2) \\
   = \dfrac{n}{2}(2 + 2n - 2) = \dfrac{n}{2}(2n) = {n^2} \\
 $
And therefore the arithmetic mean will be $\dfrac{{{\text{sum of n numbers}}}}{n}$$ = \dfrac{{{n^2}}}{n} = n$

Hence option A is correct.

Note:
Basically if we take $n = 3$ then we get the numbers as $1, 3, 5$ and their mean will be given as $\dfrac{{1 + 3 + 5}}{3} = 3$ which is the same number as we took the value of $n = 3$.