Question

The arithmetic mean of $7$ consecutive integers starting with $a$ is $m$. Then, the arithmetic mean of $11$ consecutive integers starting with $a + 2$ is
A) $2a$
B) $2m$
C) $a + 4$
D) $m + 4$

Answer 1

Hint: In order to find the arithmetic mean of the consecutive numbers, we need to know about consecutive numbers. Consecutive numbers are the numbers that have a common difference of $1$. Consecutive numbers can be found by adding one to the previous number, and so on till we want.

Complete step by step answer:
We are given with $7$ consecutive integers starting from $a$. And, we know that consecutive number means the number just after that number or we can say adding 1 to the previous number to get the consecutive number.
So, according to this if $a$ is the first number, then the second consecutive number is $a + 1$, similarly, the next consecutive number will be $a + 1 + 1 = a + 2$, and so on till we want, we can add $1$.
Therefore, we have 7 consecutive numbers, So the consecutive numbers list is:
$a,a + 1,a + 2,a + 3,a + 4,a + 5,a + 6$
Given the mean of these $7$ consecutive numbers is $m$.
Mean $ = \dfrac{{\left( {{\text{Sum of all numbers}}} \right)}}{{{\text{No of observations}}}}$
Substituting the numbers:
$ \Rightarrow m = \dfrac{{\left( {a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6} \right)}}{7}$
On solving the braces, we get:
$ \Rightarrow m = \dfrac{{\left( {7a + 21} \right)}}{7}$
$ \Rightarrow m = \dfrac{{7\left( {a + 3} \right)}}{7}$
$ \Rightarrow m = a + 3$ ……..(1)
Now, we need to find the arithmetic mean of $11$ consecutive integers starting with $a + 2$:
Therefore, adding one to the numbers, the list of $11$ consecutive integers are:
$a + 2,a + 3,a + 4,a + 5,a + 6,a + 7,a + 8,a + 9,a + 10,a + 11,a + 12$
The mean of these numbers is:
$Mean = \dfrac{{\left( {{\text{Sum of all numbers}}} \right)}}{{{\text{No of observations}}}}$
Substituting the numbers:
$ \text{Mean } = \dfrac{{\left( a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8 + a + 9 + a + 10 + a + 11 + a + 12 \right)}}{{11}}$
On solving the braces, we get:
$ \text{Mean } = \dfrac{{\left( {11a + 77} \right)}}{{11}}$
On simplification,
$ \text{Mean } = \dfrac{{11\left( {a + 7} \right)}}{{11}}$
$\Rightarrow \text{Mean } = a + 7$
Writing $a + 7$ as $\left( {a + 3} \right) + 4$ in the above equation, we get:
$ \text{Mean } = \left( {a + 3} \right) + 4$
Substituting $a + 3 = m$ from equation 1 in the above equation and we get:
$ \text{Mean } = m + 4$
Therefore, the arithmetic mean of $11$ consecutive integers starting with $a + 2$ is $m + 4$. Hence, Option (D) is correct.

Note:
Arithmetic Mean is basically the average of the numbers, which is calculated by adding all the numbers or we can say the sum of all observations divided by the number of observations.