1. 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?
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By: anil on 05 May 2019 01.51 am
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
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