Answer: B
The average of \(n\) natural numbers is: \[ \cfrac{\sum n}{n} = \cfrac{\cfrac{n(n+1)}{2}}{n} \] \[ = \cfrac{n+1}{2} \] \(\therefore\) \[ \cfrac{n+1}{2} = \cfrac{n+8}{4} \] \[ 4n + 4 = 2n + 16 \] \[ 4n - 2n = 16 - 4 \] \[ 2n = 12 \] \[ n = 6 \]
The average of \(n\) natural numbers is: \[ \cfrac{\sum n}{n} = \cfrac{\cfrac{n(n+1)}{2}}{n} \] \[ = \cfrac{n+1}{2} \] \(\therefore\) \[ \cfrac{n+1}{2} = \cfrac{n+8}{4} \] \[ 4n + 4 = 2n + 16 \] \[ 4n - 2n = 16 - 4 \] \[ 2n = 12 \] \[ n = 6 \]