Solution:
we have been asked to find the average of the first n positive even numbers.
Let the firm n positive even numbers be
[tex] 2,4,6,8,10,12,.........2(n-1), 2n [/tex]
As we know that average of numbers[tex] =\frac{\text{Sum of Numbers }}{Total number of Numbers}\\ [/tex]
Required Average[tex] =\frac{2+4+6+2(n-1)+2n}{n}\\ [/tex]
Required Average[tex] =\frac{2(1+2+3+....+(n-1)+n)}{n} [/tex]
As we know the sum of first n terms[tex] =\frac{n(n+1)}{2} [/tex]
Required Average[tex] =\frac{2n(n+1)}{2n}=n+1 \\ [/tex]
