Question

The Arithmetic mean of the set of variable \[a,a + d,a + 2d + a + 3d............a + 2nd\] is
\[\begin{array}{l}A.\,\,a + nd\\B.\,\,a - nd\\C.\,\,\left( {a + n} \right)d\\D.\,\,ad + n\end{array}\]

Answer 1

Hint: Here, arithmetic mean is also considered as mean or average. So, one can find it by using formula, sum of the terms by total number of terms to get the answer.

Complete step-by-step answer:
In the question, a set of variables are given \[a,a + d,a + 2d,a + 3d,{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. a + 2nd}}\] and we have to find arithmetic mean of it.
The arithmetic mean or mean or average is denoted as \[\bar x\] which is the mean of data set \[{x_1},{x_2},{x_3},{x_4},{x_5},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{{\rm{x}}_{\rm{n}}}\].
The arithmetic mean is the most commonly used and readily understood measure of central tendency in a data set. In statistics, the term average refers to any of the measures of central tendency. The arithmetic mean of a set of observed data is defined as being equal to the sum of numeric values of each and every value of each and every observation divided by total number of observations. Symbolically, if we have a data set consisting of the values \[{a_1},{a_2},{a_3},{a_4},{a_5},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{{\rm{a}}_{\rm{n}}}\] then, the arithmetic mean A is,
\[A = \dfrac{{{a_1} + {a_2} + {a_3} + {a_4}{\rm{ + }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{{\rm{a}}_{\rm{n}}}}}{n}\]
For example, consider the monthly salary of 10 employees of a firm: 2500, 2700, 2400, 2300, 2550, 2650, 2750, 2450, 2600, 2400.
The arithmetic mean is,
\[\dfrac{{\left\{ {{\rm{245}}0 + {\rm{26}}00 + {\rm{25}}00{\rm{ + 27}}00 + {\rm{24}}00 + {\rm{23}}00 + {\rm{255}}0 + {\rm{265}}0 + {\rm{275}}0 + {\rm{24}}00} \right\}}}{{10}} = \dfrac{{25300}}{{10}} = 2530\]
Now, in the question, data set is given: \[a,a + d,a + 2d,a + 3d,{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. a + 2nd}}\]
Here, in the data set there are \[\left( {2n + 1} \right)\] terms.
The sum of the terms are,
\[a + a + d + a + 2d + {\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. a + }}\left( {{\rm{2nd}}} \right)\]
Which can be written as,
\[\left( {2n + 1} \right)a + \left( {1 + 2 + 3 + .........2n} \right)d\]
Now, here we will apply formula that, sum of K- consecutive terms can be equal to \[\dfrac{{k\left( {k + 1} \right)}}{2}\]
So, \[1 + 2 + 3 + .....2n = \dfrac{{2n\left( {2n + 1} \right)}}{2}\]
So, the sum of terms is \[\left( {2n + 1} \right)a + \dfrac{{2n\left( {2n + 1} \right)d}}{2}\]
\[ \Rightarrow \left( {2n + 1} \right)a + n\left( {2n + 1} \right)d\]

The arithmetic mean will be calculated by dividing the sum of the terms by total number of terms.
So, we get,
\[\begin{array}{l}{\rm{Arithmetic \space mean = }}\dfrac{{\left( {2n + 1} \right)a + n\left( {2n + 1} \right)d}}{{\left( {2n + 1} \right)}}\\ \Rightarrow a + nd\end{array}\]
So, the correct option is A.

Note: The arithmetic mean generally depends on two factors: sum of the total terms and total number of terms. While calculating each of them, students must be careful about it otherwise the whole problem might get wrong.