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