Question

If the mean of the set of numbers ${x_1},{x_2},...,{x_n}$ is $\bar x$, then the mean of the numbers ${x_i} + 2i,1 \le i \le n$ is
A. $\bar x + 2n$
B. $\bar x + n + 1$
C. $\bar x + 3n$
D. $\bar x + n$

Answer 1

Hint: Here mean of the set of some numbers is given and mean of some numbers related to the given numbers is asked. To solve this type of problem, you need to use the formula of finding the mean of numbers. Basically, the mean is nothing but the average of some given numbers. It is obtained by calculating the sum of the observations divided by the total number of given observations.

Formula Used:
Mean of the numbers ${x_1},{x_2},{x_3},...,{x_n}$ is defined by $\bar x = \dfrac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}$, where $n$ is total number of given observations.
Sum of first $n$ natural numbers is $1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$
Complete step by step solution:
The given numbers are ${x_1},{x_2},{x_3},...,{x_n}$
Mean of the numbers is the sum of observations divided by total number of given observations i.e.
$\bar x = \dfrac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}$
$ \Rightarrow {x_1} + {x_2} + {x_3} + ... + {x_n} = n\bar x - - - - - \left( i \right)$
We have to find the arithmetic mean of the numbers ${x_i} + 2i,1 \le i \le n$
Sum of the numbers is $\sum\limits_{i = 1}^n {\left( {{x_i} + 2i} \right)} $
Expand the series putting $i = 1,2,3,...,n$
$ = \left( {{x_1} + 2 \cdot 1} \right) + \left( {{x_2} + 2 \cdot 2} \right) + \left( {{x_3} + 2 \cdot 3} \right) + ... + \left( {{x_n} + 2 \cdot n} \right)$
Arrange the terms.
$ = \left( {{x_1} + {x_2} + {x_3} + ... + {x_n}} \right) + 2\left( {1 + 2 + 3 + ... + n} \right)$
There are total $n$ observations required.
So, $1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$

and substitute ${x_1} + {x_2} + {x_3} + ... + {x_n} = n\bar x$ from equation $\left( i \right)$
Thus, the sum becomes $n\bar x + 2\left\{ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right\} = n\bar x + n\left( {n + 1} \right) = n\left( {\bar x + n + 1} \right)$
Dividing the sum by $n$, we get
Mean of the numbers $ = \dfrac{{n\left( {\bar x + n + 1} \right)}}{n} = \bar x + n + 1$

Option ‘B’ is correct

Note: Mean is simply average of the given numbers. If we add or subtract a number from each of the given numbers then, the mean of the resulting numbers will be equal to the mean of the given numbers with added or subtracted the number which we previously added or subtracted.