The average of 6 consecutive natural numbers is K. If the next two natural numbers are also included, how much more than K will the average of these 8 numbers be?

Let the 6 consecutive numbers be a-3,a-2,a-1,a,a+1,a+2

average = $$\frac{SumofElements}{NumberofElements}$$

It is given that average of 6 consecutive numbers be k and hence

k = $$\frac{a-3+a-2+a-1+a+a+1+a+2}{6}$$ = $$\frac{6a-3}{6}$$ = a - $$\frac{1}{2}$$

now next two numbers (a+3, a+4) are also added

Sum of 8 numbers = a-3+a-2+a-1+a+a+1+a+2+a+3+a+4 = 8a +4

average of 8 numbers = $$\frac{8a+4}{8}$$ = a + $$\frac{1}{2}$$ = k + 1

so average of 8 numbers is more than average of 6 numbers by = k+1 - K = 1