Asked by dil
                find the average of all odd natural numbers , well as that of all the even natural numbers , less than 20 compare the two averages
            
            
        Answers
                    Answered by
            oobleck
            
    well, geez, one way is just to find the average of each set:
(1+3+5+7+9+11+13+15+17+19)/10 = ____
(2+4+6+8+10+12+14+16+18)/9 = _____
Now just do your comparison.
a more elegant way, also suitable for a much larger set of values, makes use of the fact that the sum of the first n odd numbers is n^2.
So, the sum of the first 10 odd numbers is 10^2
Since even numbers are all multiples of 2, the sum of the first n even numbers is two times the sum of the numbers 1..n
Now , the sum of the numbers from 1 to n is n(n+1)/2
So, the sum of the first 9 even numbers is 9*10
    
(1+3+5+7+9+11+13+15+17+19)/10 = ____
(2+4+6+8+10+12+14+16+18)/9 = _____
Now just do your comparison.
a more elegant way, also suitable for a much larger set of values, makes use of the fact that the sum of the first n odd numbers is n^2.
So, the sum of the first 10 odd numbers is 10^2
Since even numbers are all multiples of 2, the sum of the first n even numbers is two times the sum of the numbers 1..n
Now , the sum of the numbers from 1 to n is n(n+1)/2
So, the sum of the first 9 even numbers is 9*10
                    Answered by
            ammaa
            
    really good smart work
    
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