Question

The arithmetic mean of 1,2,3, …., n, is
(a) $\dfrac{n-1}{2}$
(b) $\dfrac{n+1}{2}$
(c) $\dfrac{n}{2}$
(d) $\dfrac{n}{2}+1$

Answer 1

Hint: Here we may calculate the arithmetic mean by adding or taking the sum of all the numbers from 1 to n and then dividing the sum by n. We may calculate the sum upto n by using the formula to find the sum of terms of an AP.

Complete step-by-step answer:
We know that formula to find the average of given certain numbers is given as:

 $Avg=\dfrac{sum\,\,of\,\,numbers}{total\,\,number\,\,of\,numbers}$

So, according to this formula the average of the given numbers can be written as:
$Avg=\dfrac{1+2+3+4+......+n}{n}$

Now for finding the sum 1+2+3…..+n we may proceed as follows:

If we write the numbers as a sequence that is: 1, 2, 3, 4, ………, n.

Here, the difference between each term is 1 which is a constant. So, we can say that the given sequence is an AP with the first term equal to 1 and the common difference also equal to 1.

We know that the formula to find the sum of n terms of an AP with first term equal to a and common difference equal to d is given as:

${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$

So, using this formula to find the sum of the AP we have, the sum will be:

$\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left\{ 2\times 1+\left( n-1 \right)1 \right\} \\
 & \,\,\,\,\,\,=\dfrac{n}{2}\left( 2+n-1 \right) \\
 & \,\,\,\,\,\,=\dfrac{n}{2}\left( n+1 \right) \\
\end{align}$

Therefore, the mean will be:

$\begin{align}
  & Avg=\dfrac{\dfrac{n}{2}\left( n+1 \right)}{n} \\
 & \,\,\,\,\,\,\,\,\,\,\,=\dfrac{n+1}{2} \\
\end{align}$

Hence, option (b) is the correct answer.

Note: Students should note here that we can also use the formula ${{S}_{n}}=\dfrac{n}{2}\left( first\,term+last\,term \right)$ to find the sum of the AP so formed. Since, in this question both the first term and the last term is known, using this formula may be a shortcut to find the sum of the AP so formed and hence finding the average.