What is the average of the first n positive even numbers?

Question

asked Feb 28, 2019 122k views

2 Answers

Adam Eros · Answer 1 · 2019-03-02T15:10:54+0000

Solution:

we have been asked to find the average of the first n positive even numbers.

Let the firm n positive even numbers be

2,4,6,8,10,12,.........2(n-1), 2n

As we know that average of numbers
$=\frac{\text{Sum of Numbers }}{Total number of Numbers}\\$

Required Average
$=(2+4+6+2(n-1)+2n)/(n)\\$

Required Average
=(2(1+2+3+....+(n-1)+n))/(n)

As we know the sum of first n terms
=(n(n+1))/(2)

Required Average
$=(2n(n+1))/(2n)=n+1 \\$

Miq · Answer 2 · 2019-03-06T11:59:44+0000

The average of all of them is the same as the average of the first and last, the same as the middle number in the group:
1 + n