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

The correct option is D

$m+4$

Explanation for the correct option:

The arithmetic mean of $7$ consecutive integers starting with $a$ is $m$

common difference=$1$

That is,

$$\begin{gathered} \frac{a+\left(a+1\right)+\left(a+2\right)+\left(a+3\right)+\left(a+4\right)+\left(a+5\right)+\left(a+6\right)}{7}=m \\ \Rightarrow \frac{7a+21}{7}=m \\ \Rightarrow \frac{7\left(a+3\right)}{7}=m \\ \Rightarrow a+3=m \end{gathered}$$

Finding the arithmetic mean of $11$ consecutive integers starting with $a+2$:

$$\begin{gathered} \frac{\left(a+2\right)+\left(a+3\right)+\left(a+4\right)......+\left(a+12\right)}{11}=\frac{11a+77}{11} \\ =\frac{11\left(a+7\right)}{11} \\ =a+7 \\ =a+3+4 \\ =m+4\left[\because a+3=m\right] \end{gathered}$$

Hence, option (D) is the correct answer.