Question

The AM of \[1,3,5,...,(2n - 1)\] is
$A)n + 1$
$B)n + 2$
$C){n^2}$
$D)n$

Answer 1

Hint: First, we will see about the Arithmetic mean that is mentioned in the given problem.
The arithmetic mean is given by the sum of the observations divided by the number of observations.
Then we need to find the required sum of the given sequence and the number of terms in it to get the required answer.
Formula used: $AM = \dfrac{S}{n}$, where AM is the required arithmetic mean, S is the sum of the given sequence observations and n is the number of given sequence observations.

Complete step by step answer:
From the given, the sequence is given as \[1,3,5,...,(2n - 1)\]
First, we will calculate the total number of terms which is n, since the AM has several terms starting with one, three, five, and finishing with $2n - 1$, so these are an odd number.
So, the fourth term of the arithmetic mean is seven, the fifth term is nine hence the given sequence is an odd number.
It is also finite because we know the endpoint, that is $2n - 1$
Hence the number of in an odd integer is $n$ (because if we substitute n as any number, we get only the odd integer is given $2n - 1$)
Therefore, the number of terms in the given sequence is $n$(or we can simply apply the number of terms is n, from the given formula)
Now to calculate the sum of the terms from the given sequence \[1,3,5,...,(2n - 1)\]
From the sum of the terms in AM, this can be generalized into a term $\dfrac{n}{2}[{1^{st}}term + last term]$
Thus, the first term of the sequence is given as one and the last term of the sequence is given as$2n - 1$.
Therefore, applying the values in the generalized form of AM we get, $\dfrac{n}{2}[{1^{st}}term + last term] \Rightarrow \dfrac{n}{2}[1 + (2n - 1)]$
Further solving this we get, \[\dfrac{n}{2}[{1^{st}}term + last\, term] \Rightarrow n[n]\]
Therefore, we get the number of the AM is \[\dfrac{n}{2}[{1^{st}}term + last term] \Rightarrow {n^2}\]
Hence, we get the number of terms and sum of the terms for the given AM.
Appling thus in the given formula we get, $AM = \dfrac{S}{n} \Rightarrow \dfrac{{{n^2}}}{n} = n$
Hence, The AM \[1,3,5,...,(2n - 1)\]is $n$

So, the correct answer is “Option D”.

Note: the given sequence is an odd integer\[1,3,5,...,(2n - 1)\], we got the AM as $n$
For the same procedure to get the even integers AM as follows \[2,4,6,....,2n\](even sequence)
Thus, Applying the formula of $AM = \dfrac{S}{n}$, here the sum of the sequence is \[\dfrac{n}{2}[{1^{st}}term + last \,term] = \dfrac{n}{2}[2 + 2n] \Rightarrow n(n + 1)\]
Thus, we get $AM = \dfrac{S}{n} \Rightarrow \dfrac{{n(n + 1)}}{n} = n + 1$