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: In this question, we first need to find the \[{{n}^{th}}\] term of the given series by taking the term \[2n-1\] as some \[{{k}^{th}}\] term by using the formula \[{{T}_{n}}=a+\left( n-1 \right)d\]. Now, we need to find the sum of n terms of that series using the formula \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]. Then to get the arithmetic mean we need to divide the sum of n terms with the number of terms and simplify further.

Complete step-by-step solution-
Now, from the given series in the question we have
\[1,3,5,........\left( 2n-1 \right)\]
Here, the first term of the series is given by
\[a=1\]
Now, the common difference between consecutive terms in the series is
\[d=3-1=2\]
Let us now assume that the given term in the series \[2n-1\] as some \[{{k}^{th}}\] term
As we already know that the formula for \[{{n}^{th}}\] term of an arithmetic progression is given by
\[\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d\]
Now, we need to calculate the \[{{k}^{th}}\] term by substituting respective values and equating accordingly.
\[\Rightarrow {{T}_{k}}=a+\left( k-1 \right)d\]
Now, on substituting the respective values we get,
\[\Rightarrow 2n-1=1+\left( k-1 \right)2\]
Now, on further simplification of the above equation we get,
\[\Rightarrow 2n-1=1+2k-2\]
Now, on rearranging the terms we get,
\[\Rightarrow 2n=1+2k-1\]
Let us now simplify further and divide both sides with 2
\[\therefore k=n\]
Thus, the term \[2n-1\]is the \[{{n}^{th}}\] term of the given series.
Now, let us calculate the sum of n terms of the given series
As we already know that sum of n terms of an arithmetic progression is given by the formula
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
Now, on substituting the respective values in the above formula we get,
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2\times 1+\left( n-1 \right)2 \right)\]
Now, this can be further written in the simplified form as
 \[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2+2n-2 \right)\]
Now, on further simplification we get,
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2n \right)\]
Let us now cancel out the common terms
\[\therefore {{S}_{n}}={{n}^{2}}\]
As we already know that we get the arithmetic mean of the series on dividing the sum of n terms of the series with the number of terms in the series
\[\Rightarrow A.M=\dfrac{{{S}_{n}}}{n}\]
Now, on substituting the respective values in the above formula we get,
\[\Rightarrow A.M=\dfrac{{{n}^{2}}}{n}\]
Now, on cancelling out the common terms and simplifying further we get,
\[\therefore A.M=n\]
Hence, the correct option is (a).

Note: Instead of assuming the term \[2n-1\] as \[{{k}^{th}}\] term and then solving for the value of k we can also simplify it by directly finding the \[{{n}^{th}}\] term of the given series as we already know the first term and the common difference and simplify further. This method will have a lesser number of steps.
It is important to note that we need to divide the sum of n terms with the number of terms but not the \[{{n}^{th}}\] term. Because doing it the other way does not give the arithmetic mean.
It is also to be noted that while simplifying and substituting the values we need to consider the respective values and simplify further.