A) \[2a\]
B) \[2m\]
C) \[a+4\]
D) \[m+4\]
E) \[a+m+2\]
Correct Answer: D
Solution :
\[\because \]\[\frac{\begin{align} & a+(a+1)+(a+2)+(a+3)+(a+4) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(a+5)+(a+6) \\ \end{align}}{7}\] \[=m\] \[\Rightarrow \] \[\frac{7a+21}{7}=m\] \[\Rightarrow \] \[a=\frac{7m-21}{7}\] \[\therefore \]\[(a+2)+(a+3)+(a+4)+(a+5)\]\[+(a+6)+(a+7)\] \[\begin{align} & \underline{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,+(a+8)+(a+9)+(a+10)+(a+11) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(a+12) \\ \end{align}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11 \\ \end{align}\] \[=\frac{11a+77}{11}\] \[=\frac{11\left( \frac{7m-21}{7} \right)+77}{11}\] \[=\frac{11(m-3)+77}{11}\] \[=\frac{11m-33+77}{11}=\frac{11m+44}{11}\] \[=m+4\]
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