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 $
• E. $ a+m+2 $

Answer 1

Answer: (D)

$ \because $ $ \frac{\begin{align} & a+(a+1)+(a+2)+(a+3)+(a+4) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(a+5)+(a+6) \\ \end{align}}{7} $
$=m $
$ \Rightarrow $ $ \frac{7a+21}{7}=m $
$ \Rightarrow $ $ a=\frac{7m-21}{7} $
$ \therefore $ $ (a+2)+(a+3)+(a+4)+(a+5) $ $ +(a+6)+(a+7) $ $ \begin{align} & \underline{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,+(a+8)+(a+9)+(a+10)+(a+11) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(a+12) \\ \end{align}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11 \\ \end{align} $
$=\frac{11a+77}{11} $
$=\frac{11\left( \frac{7m-21}{7} \right)+77}{11} $
$=\frac{11(m-3)+77}{11} $
$=\frac{11m-33+77}{11}=\frac{11m+44}{11} $
$=m+4 $