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)

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